Paragraph Styles 13. A cliff diver stands at the top of a cliff and falls forwar
ID: 1877163 • Letter: P
Question
Paragraph Styles 13. A cliff diver stands at the top of a cliff and falls forward off the cliff If the cliff is 25 m tall, how fast (in m/s) is the diver going as she enters the water below? 14. A ball rolls along flat ground at 70 m/s (call this the positive direction) then encounters a hill If the ball roll: 15 m up the hill before coming to a stop, what was the acceleration (Gin m/s) experienced by the ball as it rolled up the hill? 15. While playing sock baseball in the living room, a child swings a pair of socks attached to the end of a string in a circular path above his head. If the socks travel at a constant speed of 3.5 m/s and hat is the acceleration (in m/s) experienced by the socks? the string is 0.50 m long, w accelerations of 2.5 g's at the end of a 3.5 m pole. What speed (in m/s) do they experience? to 1.9g's, but the speed should remain the same. What length of pole should be used? 16. Passengers in a whirly-go-round (a "puke your guts out" amusement ride) are subjected to 17. An engineer rides the ride in the problem above and decides the acceleration needs to be reduced 18. A ball whirls around on the end of a string, moving in a circle at a constant speed of 3.0 m/s. Ifit 19. If the string in the previous problem is replaced by a string 1.75 m long but the acceleration 20. Healthy adult humans have been shown to be able to detect accelerations as low as 10 mm's (yes experiences an acceleration of 3 m/s', how long (in m) is the string? remains unchanged, what is the ball's new speed (in m/s)? that's mm, not m) Given an airplane in flight at a cruising speed of 250 m's, what is the minimum tuming radius (in km) that will go unnoticed by the passengers?Explanation / Answer
13.
Given height of the cliff h = 25 m
The speed at the bottom is v = sqrt(2 * g * h)
v = sqrt(2 * 9.8 * 25)
v = 22.13 m/s
14. Here initial velocity u = 7 m/s
final velocity v = 0 m/s
displacement s = 15 m
from the equation v^2 - u^2 = 2 * a * s
0 - 7^2 = 2 *a * 15
a = - 1.63 m/s^2
15.
Given v = 3.5 m/s
radius of the path r = 0.5 m
acceleration a = v^2 / r
a = (3.5)^2 / 0.5
a = 24.5 m/s^2
16.
Here centripetal acceleration a = 2.5 * g = 2.5 * 9.8
a = 24.5 m/s^2
radius r = 3.5 m
from a = V^2 / r
24.5 = v^2 / 3.5
v = 9.26 m/s
17.
a = 1.9 * g = 1.9 * 9.8 = 18.62 m/s^2
v = 9.26 m/s
from a = V^2 / r
18.62 = (9.26)^2 / r
length r = 4.6 m
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