Answer:
because he's fast and speedy
Explanation:
The frequency of the wave has not changed.
In fact, the frequency of a wave is given by:

where v is the wave's speed and
is the wavelength.
Applying the formula:
- In air, the frequency of the wave is:

- underwater, the frequency of the wave is:

So, the frequency has not changed.
Answer:
w= p∆v 50000 ( 0.55-0.40) and calculate and you get it
Answer:
Torque,
Explanation:
Given that,
The loop is positioned at an angle of 30 degrees.
Current in the loop, I = 0.5 A
The magnitude of the magnetic field is 0.300 T, B = 0.3 T
We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

Let us assume that, 
is the angle between normal and the magnetic field, 
Torque is given by :

So, the net torque about the vertical axis is
. Hence, this is the required solution.