The box is in equilibrium, so Newton's second law says
<em>n</em> + (-<em>w</em>) = 0
65 N + (-<em>f</em> ) = 0
where <em>n</em> denotes the magnitude of the normal force, <em>w</em> denotes the weight of the box, and <em>f</em> denotes the magnitude of the friction force.
The box has a weight of
<em>w</em> = (25 kg) (9.80 m/s²) = 245 N
so <em>n</em> = 245 N, too.
The friction force has magnitude
<em>f</em> = 65 N
and is proportional to the normal force by a factor of <em>µ</em>, the coefficient of kinetic friction. So we have
65 N = <em>µ</em> (245 N) → <em>µ</em> ≈ 0.26
Answer:
ANGLE is 35.3 degree celcius
Explanation:
Given data:
mass m and 3m
initial speed Vi
particle with mass m is moving toward left while particle with mass 3m is moving toward right
By using conservation of momentum :
conservation of energy :
After collision, particle with mass m moves at right angles, thus by considering conservation of momentum in x & y direction,
x direction : 
y direction : 
subsitute these value in energy conservation
35.3 degree from x-axis
I have no clue sorry maybe try applying the answers and questions then I'll answer for u
100N to the left. Newton's 3rd law action and reaction
