A solid metal ball and a hollow plastic ball of the same external radius are released from rest in a large vacuum chamber. When
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Answer:
The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>
Explanation:
Given:
Upward direction is positive. So, downward direction is negative.
Tota time the ball remains in air (t) = 8.0 s
Net displacement of the ball (S) = Final position - Initial position = 0 m
Acceleration of the ball is due to gravity. So, (Acting down)
Now, let the initial velocity be 'u' m/s.
From Newton's equation of motion, we have:
Plug in the given values and solve for 'u'. This gives,
Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K =
where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a)
where
K = propagation vector
(b) The length of the string is given by:
(c) Now, we first find the frequency of the wave:
Now,
Speed of the wave is given by: