Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
An electric power measure the rate of electrical energy transfer by an electric circuit per unit of time.
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:

