The acceleration of the crate after it begins to move is 0.5 m/s²
We'll begin by calculating the the frictional force
Mass (m) = 50 Kg
Coefficient of kinetic friction (μ) = 0.15
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (N) = mg = 50 × 10 = 500 N
<h3>Frictional force (Fբ) =?</h3>
Fբ = μN
Fբ = 0.15 × 500
<h3>Fբ = 75 N</h3>
- Next, we shall determine the net force acting on the crate
Frictional force (Fբ) = 75 N
Force (F) = 100 N
<h3>Net force (Fₙ) =?</h3>
Fₙ = F – Fբ
Fₙ = 100 – 75
<h3>Fₙ = 25 N</h3>
- Finally, we shall determine the acceleration of the crate
Mass (m) = 50 Kg
Net force (Fₙ) = 25 N
<h3>Acceleration (a) =?</h3>
a = Fₙ / m
a = 25 / 50
<h3>a = 0.5 m/s²</h3>
Therefore, the acceleration of the crate is 0.5 m/s²
Learn more on friction: brainly.com/question/364384
B. Transverse Wave this is the correct answer
Winter
It is winter because the Northern Hemisphere is not facing the sun.
Clearly visible data points and appropriate labels on each access that include units
Answer:
The ratio is
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is 
Generally from Kepler's third law

Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that 
=> ![[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%5E2%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E3)
Here
is the period of Deimos
and
is the period of Phobos
So
![[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BT_1%7D%7BT_2%7D%20%5D%20%3D%20%20%5B%5Cfrac%7BR_1%7D%7BR_2%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D)
=> ![\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]](https://tex.z-dn.net/?f=%5Cfrac%7BT_1%7D%7BT_2%7D%20%20%3D%20%20%5B%5Cfrac%7B23500%20%7D%7B9380%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D)
=>