The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N,
= 25 N, a = -0.9
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.

= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
A)planets
b)the sun
c)moons
e)comets
f)asteroids
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed


The force F₁is constant acceleration is also a constant.

We need to calculate the acceleration
Using formula of acceleration



Similarly,

For total force,


The speed of second tugboat is


We need to calculate total acceleration



We need to calculate the acceleration a₂



We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂



Hence, The magnitude of F₁ is 3.7 times of F₂
I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and