Answer:
If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>
Explanation:
If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.
<u>We can prove this by the equation of heat for the two bodies:</u>
<em>According to given condition,</em>


<em>when there is no heat loss from the system of two bodies then </em>


- Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.
The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:
- when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.
OR
- the mass of colder object is half the mass of the hotter object while their specific heat is same.
A region around a charged partical or object. Let me know if this works. Hope I could help you.
He should a step-up transformer with k=220/120=1.83 so output coil must have 240*1.83=440 turns
Answer:
a)
, b) 
Explanation:
a) The potential energy is:



b) Maximum final speed:

The final speed is:


Answer:
The slope of the graph is what you need. That tells you the speed not the velocity. In order to find the velocity you would also need to know the direction of the motion.