45. 3. (III) A 7.26 kg bowling ball hangs from the end of a 2.5 m rope. The ball is pulled back until the rope makes an angle of
1 answer:
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Explanation:
1.
We use the equation
h = , where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
Answer:
2.67 m
Explanation:
k = Spring constant = 1.5 N/m
g = Acceleration due to gravity = 9.81 m/s²
l = Unstretched length
Frequency of SHM motion is given by
Frequency of pendulum is given by
Given in the question
The unstretched length of the spring is 2.67 m