Answer:
The potential energy has a maximum when the ball is a time that is half of the time for total travel
Explanation:
Generally potential energy is a the varies directly with the height according to this formula

and the ball attains a maximum height when the time is equal to half of the total time taken to travel
The main cause of this is Friction. The more oil that is laid down, the less friction there is between the ball and the lane surface. The less friction, the harder it is for the bowler to send the ball in a curved path imparted by the spin that the bowler puts on the ball at the instant of release.
Explanation:
Given that,
Rate of cooling of air
Initial temperature= 80°C
Final temperature = 5°C
We need to calculate
Using newton's law of cooling


Where, 
Here,
= 25°C (surrounding temperature)
dt = 1 minute

Put the value into the formula



Hence, This is the required answer.
Answer: 
Explanation:
Given
Current in the first wire 
Current in the second wire 
wires are
apart
Force per unit length between the current-carrying wires is

Force exerted by the wires is the same
Put the values

This force will be repulsive in nature as the current is flowing opposite
Answer:
Explanation:
Not sure what your options are but anything that says something like
"at the block surface in contact with the ramp along the line from V to Z" is probably a good shot.