2 years ago

A motor requires 400 joules of energy to lift a 5.0 kg mass 2.0 meters. Calculate the efficiency of this motor.

Physics

1 answer:

Sauron [17]2 years ago

4 0

Answer:

25%

Explanation:

use F=mg

then use the answer you get from that and plug it into W=Fxh

take that answer and divide it by 400 J and multiply by 100

round to sig figs

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b) $\Delta t'=12.5\mu s$

Explanation:

a) Let's use the constant velocity equation:

$v=\frac{\Delta x}{\Delta t}$

v is the speed of the muon. 0.9*c
c is the speed of light 3*10⁸ m/s

$t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}$

$t=5.5\mu s$

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

$\Delta t'=\gamma\Delta t$

$\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

$\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t$

$\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t$

$\Delta t'=\frac{1}{0.44}\Delta t$

No we use the value of Δt calculated in a)

$\Delta t'=2.27*5.5=12.5\mu s$

I hope it helps you!

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