A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0° below the horizontal. The coeffici
1 answer:
You might be interested in
Answer: F = 102141N
Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by
F = m.a
whose units are:
[F] = N
[m] = kg
[a] = m/s²
Jenny's car, at the moment of the break, had acceleration:
a = 78.57 m/s²
Then, Force is
F = 1300*78.57
F = 102141 N
<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>
a) For the motion of car with uniform velocity we have , , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0
Substituting
The constant velocity of car a = 2.33 m/s
b) We have
s = 520 m, t = 223 seconds, u =0 m/s
Substituting
Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209