To me "Ma" mean in real life is either mama or Massachusttes.
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:Paper and plastic!
Explanation:
Answer:
(A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
Explanation:
Given that,
Voltage = 8.2 V
Inductor = 38 mH
Resistance = 150 Ω
Time t = 0.110 ms
The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance
We need to calculate the current
Using formula of current

Put the value into the formula



(B). We need to calculate the store energy in the inductor
Using formula of energy

Put the value into the formula


{tex]E=7.04\ \mu J[/tex]
Hence, (A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.