Answer:

Explanation:
specific gravity of oil is 

we know that
change in pressure for oil is given as

here density and h is for oil

change in pressure for WATER is given as

here density is for water and h is for water

pressure change due to both is given as


If you and the source of sound are moving apart, then the pitch (frequency) <em>you hear</em> is <em>lower</em> than the pitch (frequency) that's actually leaving the source.
It doesn't matter whether you or the source is the one moving, only that the distance between you is increasing.
Answer:
Explanation:
- The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
- v = c = λ*f
where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =
4.92*10⁷ m.
Solving for f, we get the frequency of the radio waves:
f = 6.1 Hz
- Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:

- Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:
- We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:

Answer:
heat energy is used in boiling water and to make steam at power stations
Explanation:
Answer:
A homopolar motor is a direct current electric motor with two magnetic poles, the conductors of which always cut unidirectional lines of magnetic flux by rotating a conductor around a fixed axis so that the conductor is at right angles to a static magnetic field.
Explanation: