Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Answer:
graph it like thisdo the number it has so do 14
Explanation:
so do 14 and you have it
Q = mass water x specific heat water x delta T.
<span>714,000 = mass water x specific heat water x 30.
Substitute specific heat water and solve for mass water.</span>
Answer:
what do you mean by that.
