Answer:

Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )


Let north as positive
Fnet=10n-5n
=5n north

<u>Explanation:</u>
Velocity of B₁ = 4.3m/s
Velocity of B₂ = -4.3m/s
For perfectly elastic collision:, momentum is conserved

where,
m₁ = mass of Ball 1
m₂ = mass of Ball 2
v₁ = initial velocity of Ball 1
v₂ = initial velocity of ball 2
v'₁ = final velocity of ball 1
v'₂ = final velocity of ball 2
The final velocity of the balls after head on elastic collision would be

Substituting the velocities in the equation

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.
Your center or gravity helps your balance because either it will help younot fall or it wont some people are just clusmy
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