EXAM 1 October 2, 1998 PHYS 207
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g = 9.8 m/sec2 = 32 ft/sec2 |
Radius of the earth = 6.38 x 106 m | The mass of the electron is 9.11 x 10-31 kg. |
Calculus and trig formulas will not be given after this exam: | ||
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cos(2x) = cos2 (x) - sin2 (x) | sin(2x) = 2 sin(x) cos(x) | d xn / dx = n xn-1 |
d cos(x)/dx = - sin(x) | d sin(x)/dx = cos(x) | d(fg)/dx = f (dg/dx) + (df/dx) g |
1. (10 points)
(a) State and explain how to remember (not necessarily derive) the relations for the position and velocity of a particle as a function of time for motion in one dimension with constant acceleration.
(b) Sketch a single v-versus-t curve in which there are (labelled) points or segments where
The remaining problems count 30 points each. Include a diagram whenever possible. If you work a problem symbolically, you need not substitute numerical values into your answer.
2. (From homework) Calculate the acceleration of a person at latitude 40o owing to the rotation of the Earth. (The latitude of the equator is 0o and of the pole is 90o).
3. An elevator ascends from the ground at a speed of 0.5 m/sec. A child drops a marble through a hole in the floor, and the marble hits the ground 2.5 sec after it is dropped. Find the height of the elevator above the ground at the time the marble is released.
4. A merry-go-round whose radius is R meters is rotating at an angular speed of w radians per second. A "punkin chunker" from the contest at Lewes is placed on the outer edge of the merry-go-round and launches a pumpkin at a velocity relative to itself of v0 m/sec at an angle of 45o from the horizontal. The initial velocity is perpendicular to the line from the chunker to the axis of the merry-go-round. How far from the launch point does the pumpkin land? Show on a diagram the merry-go-round and the path of a point which stays directly beneath the pumpkin. (You may need additional diagram(s) to work the problem successfully.)