Answer:
50.8 watt
Explanation:
we know that P=W÷t
W=F.S S-->distance=50 ft= 15.24 m
F=ma
=100×10=1000 N
SO W= 1000×15.24
=15240 J
NOW
P=W÷t t=5 mints = 5×60=300 sec
P=15240÷300
P=50.8 watt
You are exerting 100N. Since there’s no NET force, then there must be exactly 100N pushing exactly back on your 100N to cancel it to exactly zero. Newton's first law states that whether a body is at rest or travelling in a straight line at a constant speed, it will remain at rest or continue to move in a straight line at a constant speed unless acted upon by a force.
The answer would be:
Doubling the height will increase the amount of Joules produced.
Answer with Explanation:
We are given that




a.Length of segment,l=20 m
Magnetic force ,F=

Substitute the values

Hence, the magnetic force exert by each segment on the other=0.0119 N
b.We know that when current carrying in the wires are in same direction then the force will attract to each other.
Hence, the force will be attractive.
The amount of heat needed to increase the temperature of a substance by

is given by

where m is the mass of the substance, Cs is its specific heat capacity and

is the increase of temperature.
If we re-arrange the formula, we get

And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance: