Given :
Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.
To Find :
Will the heat produced in both the cases be equal.
Solution :
Heat released is given by :
H = i²Rt
Here, R is resistance and is given by :

So,
Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.
So, heat produced by Reem's wire is more than Nain one.
Hence, this is the required solution.
Answer:
Your answer would be C, Radio waves.
Explanation:
Photons are particles of electromagnetic radiation.
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is
The temperature of water is
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)


Answer with Explanation:
Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.
For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference
While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.