Answer:0.502kg
Explanation:
F4om the relation
Power x time = mass x latent heat of vapourization
P.t=ML
1260 * 15 *60 = M * 22.6 * 10^5
M= 1134000/(22.6 *10^5)
M=0.502kg=502g
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "
Doppler's effect."
Now the general formula of the Doppler's effect is:

-- (A)
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.

= Velocity of the receiver/observer relative to the medium = ?.

= Velocity of the source with respect to medium = 0 m/s.

= Frequency emitted from source = 400 Hz.

= Observed frequency = 408Hz.
Plug-in the above values in the equation (A), you would get:


Solving above would give you,

= 6.8 m/s
The correct answer = 6.8m/s
The source charges' magnitude is signified by the arrows pointing outward. The more arrows there are, the greater is its magnitude. This is because, each arrow represents an electrical force exerted by the source. When you add up all the arrows there is, the electrical force becomes even greater. The answer in descending order would be C > A > B > D.
Answer:
Frequency will be equal to 5.20 kHz
So option (c) will be correct answer
Explanation:
We have given value of capacitance 
Potential difference across capacitor V = 12 volt
Current through capacitor 
Capacitive reactance will be equal to 
Capacitive reactance is equal to 




f = 5.20 kHz
So frequency will be equal to 5.20 kHz
So option (c) will be correct answer
Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
We know that


n and v remain same at both side. so we have

..............1
let final pressure is P and temp 

..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, temperature on left side is 1.48 times the temperature on right