A constant velocity implies the two forces must be equal and opposite.
Friction acts horizontal to the ground, therefore we must find the force applied to the sled rope that acts horizontal to the ground.
Do this by resolving:
Force = 80cos53
The force opposing this is equal, and so also = 80cos53 = 48 N (2 sig. fig.)
Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
A. the carbons are unbalanced
B. the hydrogens are unbalanced.
D. the chlorines are unbalanced.
That leaves C. to be correctly balanced.
Answer:
V = 3.17 m/s
Explanation:
Given
Mass of the professor m = 85.0 kg
Angle of the ramp θ = 30.0°
Length travelled L = 2.50 m
Force applied F = 600 N
Initial Speed u = 2.00 m/s
Solution
Work = Change in kinetic energy
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