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sladkih [1.3K]
2 years ago
13

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in

3.1 min, starting and ending at rest. The elevator's counterweight has a mass of only 995 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?
Physics
1 answer:
insens350 [35]2 years ago
5 0

Answer:

659.01W

Explanation:

The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be

Wc = mass of the cab × acceleration due to gravity in m/s²

Wc = 1250 × 9.81 = 12262.5 N

but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N

Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N

the motor of the elevator will have to provide this in form of work

work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m

power provided by the motor of the elevator = workdone by the motor / time in seconds

Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W

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KatRina [158]

Answer:

Option D - 0.2 s

Explanation:

We are given;

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Grop 17 of the periodic table contains the ?
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yulyashka [42]

The  force applied to lift the crate is 171 N

Explanation:

The lever works on the principle of equilibrium of moments, so we can write:

F_i d_i = F_o d_o

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d_i is the arm of the input force

F_o is the output force

d_o is the arm of the output force

For the lever in this problem, we have:

d_i = 0.25 m

d_o = 0.19 m

F_i = 130 N (force applied)

Solving the equation for F_o, we find the force applied to lift the crate:

F_o = \frac{F_i d_i}{d_o}=\frac{(130)(0.25)}{0.19}=171 N

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

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