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Gre4nikov [31]
3 years ago
7

A 500 N passenger weighs 24500 N elevator that rises 30 m in exactly 1 minute how much power is needed for the elevators trip

Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

12500 W (12.5 kW)

Explanation:

The total weight of the passenger + elevator is

F_g = 500 N + 24500 N=25000 N

The total work done to rise the elevator + passenger is equal to the product between the total weight and the distance covered during the trip (d = 30 m):

W=F_g d = (25000 N)(30 m)=750,000 J

The power needed for the trip is equal to the ratio between the work done (W) and the time taken (t):

P=\frac{W}{t}

Since the time taken is t = 1 min = 60 s, the power needed is

P=\frac{750,000 J}{60 s}=12,500 W=12.5 kW

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First we must know the travel time of the ship seen from the earth. The spaceship travels at half the speed of light, this means that the amount of time the spacecraft must spend to travel the same distance is double compared to the light, that is 60 years.

Now due to the speed of the ship, we must take into account relativistic effects, such as time dilation, this is given by:

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Rewriting for t:

t=t'\sqrt{1-\frac{v^2}{c^2}}\\t=60\sqrt{1-\frac{(0.5c)^2}{c^2}}\\t=60\sqrt{1-0.5^2}\\t=51.96 years

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3 years ago
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