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.
Work = (force) x (distance)
80 J = (force) x (4 m)
Force = (80 J) / (4 m) = 20 N
That's IF the force was in the same direction as the 4m of motion.
If the force was kind of slanted, then it had to be stronger, and
it had a component of 20N in the direction of the motion.
Answer:
<h2>18 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 6 × 3
We have the final answer as
<h3>18 N</h3>
Hope this helps you
MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s so you are left with Solution 4.78 10-19 J OR .478 aJ <span>Apex - 467 nm ^.^ hopefully thats the correct thing</span>
Your answer is.......a car moved 60 km East and 90 km in west.
