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2 years ago

Find the efficiency of a machine that does 800 J of work if the input work is 2,000 J.

Physics

1 answer:

torisob [31]2 years ago

3 0

Answer:

Efficiency: 0.4 (40%)

Explanation:

The efficiency of a simple machine is given by the ratio

$\eta = \frac{W_{out}}{W_{in}}$

where

$W_{in}$ is the input work of the machine

$W_{out}$ is the output work of the machine

For the machine in this problem, we have:

$W_{out}=800 J$ is the output work

$W_{in}=2000 J$ is the input work

Therefore, the efficiency of the machine is

$\eta=\frac{800}{2000}=0.4$

Which can be also written as percentage:

$\eta=0.4\cdot 100 = 40\%$

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2 years ago

The wavelength of green light is 550 nm.

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Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

velocity of electron is 1323.88 m/s

momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

velocity of electron is 1654.85 m/s

momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

$p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s$

velocity of electron is given by;

$P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v = \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s$