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course phy 242
5/28/13 around 3:00 pm
1)Changes 2 mph/s, we are looking for seconds or time it will take for the speedometer to move from 20 mph to 30 mph.30 mph -20 mph = 10 mph interval
In order to gain seconds from mph/s and mph, we must divide mph by mph/s which equals seconds. Now doing that with the numbers: 10 mph / (2 mph/s) = 5 seconds.
Now we are to find the mph given 2mph/s and 10 mph and the time of 7 seconds later.
x- 10 mph = y mph
y mph/ (2mph/s) = 7 seconds
y = 14 mph
x- 10 mph = 14 mph
x = 24 mph, it will read 24 mph.
2)Rate of speed = 2 mph/s
Speed = 10 mph
Time required = 10 s
Yes without any calculations one can conclude that since its passes the milepost at a speed of 20 mph and the speed is increasing by 2 mph/s then the car will require less than 10 seconds to reach the lamppost when compared to the previous data given.
No, the speed at the lamppost will not be 10 mph greater than before, because first it must be considered that the distance traveled is equal and so since that’s kept constant along with the rate of change of speed, the concentration is on the time and speed. Since the initial speed of 20 mph has lesser time than that of initial speed of 10 mph then therefore, it will have lesser time to accelerate to a speed and so therefore will be lesser than that of 10 mph.
3)Speed change = 30 mph - 20 mph = 10 mph
Time = 5 s
Rate =10 mph / 5 s= 2 mph/s
Speed change = 90 mph - 40 mph = 50 mph
Time= 20 s
Rate = 50 mph / 20 s = 5/2 mph/s = 2.5 mph/2
Therefore, the automobile that speeds up from 40 mph to 90 mph has the greatest rate.
Self-critique: I did not mention the velocity difference between 2.5 mph/s and 2 mph/s.
4)Team one
m= 1500 kg
net force= 3000 N
rate = 3000/1500 = 2 N/kg
Team two
m= 2000 kg
net force = 5000 N
rate= 5000/2000= 5/2 = 2.5 N/kg
Therefore, team two can more quickly accelerate their initially stationary automobile to 5 mph faster, due to the greater velocity rate.
Team two
Net force = 5000-500= 4500 N
m= 2000 kg
rate = 4500/2000= 2.25 N/kg
No the other team would still not win.
5)In this case, if the 200 Ib player was moving in the same velocity then the 200 Ib player would have been moving backwards immediately. In this case sine one the players mass is greater than the other and the other player’s velocity is greater than the other, it cannot be predicted without calculations.
Player 1
250 Ib * 10 ft/s= 2500 Ib*ft/s
Player 2
200 Ib * 20 ft/s = 4000 Ib*ft/s
Therefore, with calculations player 1 would move backwards.
6)In order to determine who can climb further up the mountain we must use the amount of
cereal consumed and the mass of the climbers to make a rate which we can compare.
12 oz/ 200 Ib= 3/50 oz/Ib = 0.06 oz/ Ib
10 oz /150 Ib = 1/15 oz/Ib = 0.067 oz/ Ib
Therefore the 150 Ib climber can climb further up the mountain.
7)Since the steepness of the mountain is constant, it will require both vehicles to the same rate to change their velocities; therefore it will take vehicle with twice the speed to take twice as long to stop.
Since the slope change will be the same, velocity would be changing at a constant rate; therefore, the average coasting will be twice as great for the car with twice the velocity.
The distance traveled by the vehicle with twice the speed will be 4 times the distance of the normal speed vehicle. This is due to one of the kinematics equation where velocity = 2* acceleration* distance. I here we can neglect acceleration since the slope is the same. So if we use x as the velocity of the vehicle normal speed then we get x = 2d, so if we multiply the velocity by 2 then the other side of the equation must be multiplied by 2 too so therefore we get 2x = 4d.
Self-critique: I did not mention the neglecting of air resistance for the entire question.
8)First we must find the rate of the stretching of the rope. For the first one 5ft/100Ib = 1ft/20 Ib
Next person 9ft/150 Ib = 3ft/50 Ib , next person 12ft/200 Ib = 3ft /50 Ib .
Now that it is confirmed that 3 ft /50 Ib , we divide the 125/50 = 2.5 and then multiply by the 3 ft = 7.5 ft which is greater than 7 ft.
9)Since the distance and the force is also doubled due to the behavior of the sling shot increasing force as it is pulled back, it produces a quadruple travel distance than that of the pulled back 4 feet.
10)From a distance of half a mile the minimal difference between the sphere cannot be determined, therefore the moth does not distinguish the difference between them.
Since the first glass sphere has a smaller areas due to its smaller diameter, it will distribute more light over its area and so seem brighter than that of the second sphere which can distribute lesser light over its entire area. To moth the first sphere with diameter of 1 will seem brighter. Surface area of a sphere is SA= 4*pi*r^2.
So for the one sphere with diameter 1 foot the SA is pi while the other sphere is 4pi. This demonstrates a 4 times difference between the surface areas. And so they want the ratio of 2 diameter to that of 1, therefore 4pi:pi = 4:1. Therefore the second sphere with diameter of 2 will have 1/4 times the brightness than that of the first sphere. So the second sphere will appear to have less than half the brightness of the first.
11)Increasing the temperature of the ice by 20 degrees to reach its melting point, increasing the temperature of the water by 20 degrees after all the ice melted, and melting the ice at its melting point.
Ice seems to melt at 0 degrees Celsius, because when the most of the ice is melted the temperature is still at 0 degrees Celsius.
12)If the peaks of the approaching waves are each 6 inches high then I am expected to bob up and down 6 inches + 6 inches = 12 inches or 1 foot.
One must move 12 in + 6 inch = 18 in = 1.5 ft.
1.5 ft + 1.5 ft = 3 ft move to move toward one end or the other in order for peaks to meet
valleys.
13)The steel ball would be expected to have the greater average velocity as it falls considering air resistance.
The steel ball would be expected to spend the greater time falling considering air resistance.
The steel ball will hit the ground first considering air resistance.
14)Power = Current* Voltage
Now if you double the voltage => 2*V*I= 2P, then you must double the power as well. Therefore twice the power is required if the voltage is doubled.
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Self-critique (if necessary):
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Self-critique rating:
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Very good.
I've prepared two documents to email you to show you how to submit work. However I can't connect to anything at VHCC tonight, so it will probably be tomorrow before I can send them.
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