Sarah's acceleration is 
Explanation:
The force of kinetic friction acting on Sarah has a magnitude which is given by:

where
is the coefficient of kinetic friction
m is Sarah's mass
g is the acceleration of gravity
Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

where a is the acceleration
Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

And substituting the following values:
(coefficient of friction)
(acceleration of gravity)
we find:

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ANSWER

EXPLANATION
We want to convert 12000 inches to yards.
To do this, divide the value in inches by 36:

That is the answer.
Answer:

Explanation:
<u>Accelerated Motion
</u>
When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

where a is the acceleration, and vo is the initial speed
.
The train has two different types of motion. It first starts from rest and has a constant acceleration of
for 182 seconds. Then it brakes with a constant acceleration of
until it comes to a stop. We need to find the total distance traveled.
The equation for the distance is

Our data is

Let's compute the first distance X1


Now, we find the speed at the end of the first period of time


That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0



Computing the second distance


The total distance is



Answer:
25.6 m/s
Explanation:
Draw a free body diagram of the sled. There are two forces acting on the sled:
Normal force pushing perpendicular to the hill
Weight force pulling straight down
Take sum of the forces parallel to the hill:
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 38.0°)
a = 6.03 m/s²
Given:
v₀ = 0 m/s
a = 6.03 m/s²
t = 4.24 s
Find: v
v = at + v₀
v = (6.03 m/s²) (4.24 s) + (0 m/s)
v = 25.6 m/s