Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t
Case I
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When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C
Case II
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When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C
Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C
Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C
Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
a) 1.57 m/s
The sock spins once every 2.0 seconds, so its period is
T = 2.0 s
Therefore, the angular velocity of the sock is

The linear speed of the sock is given by

where
is the angular velocity
r = 0.50 m is the radius of the circular path of the sock
Substituting, we find:

B) Faster
In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

Therefore, the new linear speed would be:

And substituting,

So, we see that the linear speed has doubled.
Answer:
Option (e)
Explanation:
If a mass attached to a spring is stretched and released, it follows a simple harmonic motion.
In simple harmonic motion, velocity of the mass will be maximum, kinetic energy is maximum and acceleration is 0 at equilibrium position (at 0 position).
At position +A, mass will have the minimum kinetic energy, zero velocity and maximum acceleration.
Therefore, Option (e) will be the answer.
Answer:
no
Explanation:
it is faster at the equator