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:
4200 Joules
Explanation:
Work done =force x distance
From the question , we’re given f =350N ,
d = 12m
Using the above formula, we have
Workdone = 350 x 12
= 4200 Joules
In Rutherford's gold foil experiment, some of the positive particles would pass through the foil and some would bounce off. This led to a new theory that all of the positive subatomic particles were in the center of the atom instead of evenly spread throughout.
Yup, I think you add all of them
<span>ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)</span>
