course Phys 201
024. `query 24
Question: **** `qWhy was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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Your solution:
It is important the string have slack due to the string having a cent. a equal to gravity so it goes in the diesired direction.
Confidence Assessment: 3
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Given Solution:
`a** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity.
If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight.
If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **
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Self-critique (if necessary): i understand this problem
Self-critique Rating:3
Question: **** `qWhy do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
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Your solution:
The top is directly center of the radian line and therfore, exactly perpendicular. That makes the velocity horizontal.
Confidence Assessment: 2
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Given Solution:
`a** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. **
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Self-critique (if necessary): I understand this problem
Self-critique Rating:3
Question: **** `qWhat is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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Your solution:
The centripetal acceleration will be equal to the acceleration of gravity.
Self-critique Rating:3
Question: **** `qQuery principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?
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Your solution:
10BLOCK=Xaxis
-2 blocks= y axis
ds= sqrt. ((10blocks)^2+(-2blocks)^2)
=sqrt. (100blocks^2+2blocks^2)
=sqrt. (104blocks^2)
=sqrt.104^2
=10.2 blocks
tan^-1(-2blocks/10blocks)
tan^-1 (-.2)
=-11.3deg SE
Confidence Assessment: 3
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Given Solution:
`aThe final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction.
The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2)
= sqrt(104 blocks^2)
= sqrt(104) * sqrt(blocks^2)
= 10.2 blocks.
The direction makes and angle of
theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees
with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east.
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Self-critique (if necessary): i understand this problem
Self-critique Rating:
Question: **** `qQuery principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land?
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Your solution:
V0=0 vert
a=9.8m/s^2
dt=3s
Vave=1.8m/s horiz.
Vert ds= Vo*dt+.5a*dt^2
=0m/s*3s+.5*9.8m/s^2*3s^2
=4.9m/s^2*9s
ds=44.1m
horiz. ds= Vave*dt
=1.8m/s*3s
ds=5.4m
Confidence Assessment: 2
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Given Solution:
`aThe diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds.
The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore
vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m.
The cliff is therefore 44 m high.
The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have
horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters.
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Self-critique (if necessary):
&#Very good work. Let me know if you have questions. &#
course Phys 201
024. `query 24
Question: **** `qWhy was it necessary to let the string go slack at the top of the circle in order to get the desired results?
*********************************************
Your solution:
It is important the string have slack due to the string having a cent. a equal to gravity so it goes in the diesired direction.
Confidence Assessment: 3
*********************************************
Given Solution:
`a** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity.
If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight.
If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **
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Self-critique (if necessary): i understand this problem
Self-critique Rating:3
Question: **** `qWhy do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
*********************************************
Your solution:
The top is directly center of the radian line and therfore, exactly perpendicular. That makes the velocity horizontal.
Confidence Assessment: 2
*********************************************
Given Solution:
`a** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity, being perpendicular to this vertical, must be entirely in the horizontal direction. **
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Self-critique (if necessary): I understand this problem
Self-critique Rating:3
Question: **** `qWhat is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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Your solution:
The centripetal acceleration will be equal to the acceleration of gravity.
Self-critique Rating:3
Question: **** `qQuery principles of physics and general college physics 7.02. Delivery truck 18 blocks north, 10 blocks east, 16 blocks south. What is the final displacement from the origin?
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Your solution:
10BLOCK=Xaxis
-2 blocks= y axis
ds= sqrt. ((10blocks)^2+(-2blocks)^2)
=sqrt. (100blocks^2+2blocks^2)
=sqrt. (104blocks^2)
=sqrt.104^2
=10.2 blocks
tan^-1(-2blocks/10blocks)
tan^-1 (-.2)
=-11.3deg SE
Confidence Assessment: 3
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Given Solution:
`aThe final position of the truck is 2 blocks south and 10 blocks east. This is equivalent to a displacement of +10 blocks in the x direction and -2 blocks in the y direction.
The distance is therefore sqrt( (10 blocks)^2 + (-2 blocks)^2 ) = sqrt( 100 blocks^2 + 4 blocks^2)
= sqrt(104 blocks^2)
= sqrt(104) * sqrt(blocks^2)
= 10.2 blocks.
The direction makes and angle of
theta = arcTan(-2 blocks / (10 blocks) ) = arcTan(-.2) = -12 degrees
with the positive x axis, as measured counterclockwise from that axis. This puts the displacement at an angle of 12 degrees in the clockwise direction from that axis, or 12 degrees south of east.
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Self-critique (if necessary): i understand this problem
Self-critique Rating:
Question: **** `qQuery principles of physics and general college physics 7.18: Diver leaves cliff traveling in the horizontal direction at 1.8 m/s, hits the water 3.0 sec later. How high is the cliff and how far from the base does he land?
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Your solution:
V0=0 vert
a=9.8m/s^2
dt=3s
Vave=1.8m/s horiz.
Vert ds= Vo*dt+.5a*dt^2
=0m/s*3s+.5*9.8m/s^2*3s^2
=4.9m/s^2*9s
ds=44.1m
horiz. ds= Vave*dt
=1.8m/s*3s
ds=5.4m
Confidence Assessment: 2
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Given Solution:
`aThe diver's initial vertical velocity is zero, since the initial velocity is horizontal. Vertical velocity is characterized by the acceleration of gravity at 9.8 m/s^2 in the downward direction. We will choose downward as the positive direction, so the vertical motion has v0 = 0, constant acceleration 9.8 m/s^2 and time interval `dt = 3.0 seconds.
The third equation of uniformly accelerated motion tells us that the vertical displacement is therefore
vertical motion: `ds = v0 `dt + .5 a `dt^2 = 0 * 3.0 sec + .5 * 9.8 m/s^2 * (3.0 sec)^2 = 0 + 44 m = 44 m.
The cliff is therefore 44 m high.
The horizontal motion is characterized 0 net force in this direction, resulting in horizontal acceleration zero. This results in uniform horizontal velocity so in the horizontal direction v0 = vf = vAve. Since v0 = 1.8 m/s, vAve = 1.8 m/s and we have
horizontal motion: `ds = vAve * `dt = 1.8 m/s * 3.0 s = 5.4 meters.
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Self-critique (if necessary):
&#This looks good. Let me know if you have any questions. &#