To determine, the final velocity we use one kinematic equation. It would be 2ax = vf^2 - v0^2. The initial velocity would be zero since it starts from rest. Therefore, the equation would be:
vf = √(2ax) where a is the acceleration ( a = g ) and x is the height
vf = √(2)(9.8)(25)
vf = - 22.14 m/s
Answer:
reddish-orrange
Explanation:
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Answer:
Explanation:
Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:
a = -9.8 m/s/s
v₀ = 0 (since the ball was held before it was dropped)
v = ??
Δx = -8 m (negative because the ball drops this far below the point from which it was released).
Putting all this together in one equation:
v² = v₀² + 2aΔx and filling in this equation:
v² = (0)² + 2(-9.8)(-8) and
v² = 156.8 so
v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)
Answer:
Part a)

Part b)

Explanation:
Force applied by the student on the box is 80 N at an angle of 25 degree
so here two components of the force on the box is given as




now in vertical direction we can use force balance for the box to find the normal force on it



now kinetic friction on the box opposite to applied force due to rough floor is given as


now the net force on the box in forward direction is given as


now the acceleration of the box is given as


Part b)
when box is pulled up along the inclined surface of angle 10 degree
now the two components of the force will be same along the inclined and perpendicular to inclined plane


now force balance perpendicular to inclined plane is given as


now the friction force opposite to the motion on the box is given as


now the net pulling force along the inclined plane is given as



now the box will decelerate and it is given as

