Answer:
the friction force in the reverse direction is 200 *0.4=80 N.
the net forward force acting on the box is therefore
Fnet= 100 - 80 N
= 20 N
acceleration = Fnet / mass
=Fnet *g/(weight)
=20 *9.8/200 = 0.98 m/s^2
Explanation:
Answer:
9.22 s
Explanation:
One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad
Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:
For the child: 
For the horse: 
For the child to catch up with the horse, they must cover the same angular distance within the same time t:



t = 25.05 or t = 9.22
Since we are looking for the shortest time we will pick t = 9.22 s
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.
Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s
Answer: 11.4 s
Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²
Answer: 44.7 m/s²
Answer:
pretty sure its B if it isnt im so so sorry
Explanation: