What power (in kw) is supplied to the starter motor of a large truck that draws 275 a of current from a 26.0 v battery hookup?
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Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.
The speed he was moving at when he finished falling is 30 m/s.
The given parameters;
mass of the bungee, m = 80 kg
impulse provided by the rope, J = 3200 Ns
initial upward velocity of the jumper, u = 10 m/s
- Let the final velocity after falling = v
- Let the upwards motion = negative
- Let the downwards motion when falling = positive
Apply the principle of conservation of linear momentum;
J = ΔP = Δmv = m(v - u)
3200 = 80(v - (-10))
3200 = 80(v + 10)
Thus, the speed he was moving at when he finished falling is 30 m/s.
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