#$&* course Mth 164 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: ** At 3 rad/sec a complete trip around the reference circle takes 2 pi / 3 seconds, close to but not exactly 2 seconds. 2 pi / 3 seconds is the distance between the peaks on the graph of y vs. t. If the circle has radius 5 the max and min will be 5 units above and below the center of the circle, at 12 - 5 = 7 and 12 + 5 = 17. **
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15:10:20
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After studying the given solution, I am still confused. Critique #2:
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15:22:13 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We position the circle with the center halfway between minimum and maximum values. We find the center by adding the min and max values and dividing by two to get the average. The radius can be found by finding the difference of the min and max and dividing difference by 2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The center of the circle will be halfway between the max and min values, which can be found by averaging the two values (i.e., add and divide by 2). The diameter will be the difference between the max and min values and the radius will be half of the diameter. **
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15:22:16 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK &&&& What is the vertical coordinate of the center of the circle, and what is the angular velocity of the reference point, for the daylight model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we compare a day to a circle, it takes 24 hrs to complete one revolution. Therefore we have: angular speed = 1 rev / 24 hrs * 2pi rad / rev = 2pi/24 rad/hrs = pi/12 rad/hrs or pi/12 hrs The difference from one day to the next is 24 hrs. Therefore 24/2 is the vertical coordinate, which is 12. Attempt#2: If the cycle equals 1 year and the length of a day varies from 9 hrs to 15 hrs over this period we can model this data on a graph of y vs t. omega = theta / t omega = (2 pi rad)/ 12 months = pi/6 rad/month If we let y = the number of hrs then the center of the circle would be (9 + 15) / 2 = 12 = y confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If the period is 52 weeks then you have 2 pi / 52 cycles in a week or pi/26 cycles per week. If the period is in months then you have 2 pi / 12 cycles per month, or pi/6 cycles per month. The vertical coordinate of the center will be the day length midway between the min and max day lengths, which is 12 hours.**
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not sure if I understand the jargon being used. ???Is a cycle the distance, with respect to time, between peaks(period T) on the graph y vs t and the wavelength on the graph y vs x. Disregard this self critique…
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Given Solution: ** If the period is 52 weeks then you have 2 pi / 52 cycles in a week or pi/26 cycles per week. The vertical coordinate of the center will be the temperature midway between the min and max temperatures.**
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15:58:33 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???I don’t understand what angular velocity of the reference point means. Disregard this self critique…
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Given Solution: ** If you have a cycle in 10 hours then you have 2 pi rad in 10 hours, or 2 pi / 10 = pi/5 rad / hour. The vertical coordinate of the center will be the water level midway between the min and max water levels. **
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15:58:35 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK &&&& What is the vertical coordinate of the center of the circle, and what is the angular velocity of the reference point, for the ocean wave model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This question depends on whether you are modeling 1 wave or several. The model for 1 wave could start at the minimum water height and end at a value equal to this. In this case, the angular speed would 2pi r/some unit of time. If we are modeling the wave height over several cycles then we would measure the height of each wave and this would determine the minimum and maximum values to be used in the model. In this case, the center of the circle is located midway between the minimum and maximum wave height. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The center will lie halfway between the highest and lowest levels. At 5 waves per minute the angular frequency would be 5 periods / minute * 2 pi rad / period = 10 pi rad / min. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&!#*&! &&&& query ch. 5 # 78 15 in wheels at 3 rev/sec. Speed in in/s and mph: rpm?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The radius of the wheels for this question is 15 in. If the wheels making 3 revolutions per second, we can find the angular speed as follows: omega = angular speed omega = 3 rev/sec = (3 rev/sec)*(2pi rad/rev) = 6pi /sec v = linear speed = r * omega v = 15 in * 6pi rad/sec = 90pi in/sec or approx. 283 in/sec v = (90pi in/sec) * (mi/63360 in) * (3600 sec/hr) = 324000pi/63360 mi/hr or approx. 16.1 mi/hr Another way to approach this question is to note that 1 revolution of the wheel occurs every 1/3 second. C = 2pi r; where r is radius and C is circumference. Using the formula we have: C=2pi*15 inches C=30pi inches 30pi inches/(1/3 second) = 90pi inches/second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If 15 inches is the diameter of the wheel then the radius is 15 inches. The angular velocity is 3 rev / sec * 2 pi rad / rev = 6 pi rad / sec. Each radian of angular displacement corresponds to a distance along the arc which is equal to the radius. So 6 pi rad / sec * 15 inches / radian = 90 pi inches / second. If you approximate this you get around 280 in/sec. This is 280 in / sec * 1 ft / 12 in = 23 ft / sec approx. A mile is 5280 ft and an hour is 3600 sec so this is 23 ft/sec * 1 mile / 5280 ft * 3600 sec / 1 hr. = 16 miles / hr approx.. ** Note that 3 revolutions / second is 180 revolutions / minute, since there are 60 seconds in a minute.
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