Answer:
impulse acting on it
Explanation:
The impulse is defined as the product between the force applied to an object (F) and the time interval during which the force is applied (
):

We can prove that this is equal to the change in momentum of the object. In fact, change in momentum is given by:

where m is the mass and
is the change in velocity. Multiplying and dividing by
, we get

and since
is equal to the acceleration, a, we have

And since the product (ma) is equal to the force, we have

which corresponds to the impulse.
Answer:
Explanation:
Given
mass of boy=36 kg
length of swing=3.5 m
Let T be the tension in the swing
At top point 
where v=velocity needed to complete circular path
Th-resold velocity is given by 

So apparent weight of boy will be zero at top when it travels with a velocity of 
To get the velocity at bottom conserve energy at Top and bottom
At top 
Energy at Bottom 
Comparing two as energy is conserved



Apparent weight at bottom is given by

Answer:
The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from 4o C to 0o C, and it becomes less dense. The density is maximum at 4 degree centigrade and decreases below that temperature as shown in graph.
Explanation:do you want me to explain it more??
"Voltage" is the "pressure" that makes electrons want to leave where they are
and head in some direction, if there's conducting material in that direction.
"Current" is the rate at which they all migrate in that direction.