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4 years ago

6

Define output work and input work

1 answer:

Olenka [21]4 years ago

6 0

Answer: Input work is the work done on a machine as the input force acts through the input distance.Other ways it would be the work done on a body or system, that is, forces that are applied to the body or system. This is in contrast to output work which is a force that is applied by the body or system to something else.Output work is the work done by a machine as the output force acts through the output distance. The machine does to the object to increase the output distance.

Explanation:

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What is the density of 25 grams

Karolina [17]

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3 years ago

Calculate the momentum of a 2500kg car with a velocity of 20 m/s

Sedaia [141]

Answer:

50000 kg.m/s

Explanation:

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4 years ago

A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic

vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 $m^{2}$ </h3>

Explanation:

Given :

No. of turns in the first coil $N_{1} = 41$

No. of turns in the second coil $N_{2} = 123$

Area of first coil $A_{1} = 0.074 m^{2}$

According to the law of electromagnetic induction,

Induced emf = $-N \frac{d \phi}{dt}$

Where $\phi =$ magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

$-\frac{N_{1} d\phi _{1} }{dt_{1} } = -\frac{N_{2} d\phi _{2} }{dt_{2} }$

$\frac{N_{1} A_{1} dB_{1} }{dt_{1} } = \frac{N_{2} A_{2} dB_{2} }{dt_{2} }$

$A_{2} = \frac{N_{1} A_{1} }{N _{2} }$

$A_{2} = \frac{41 \times 0.074 }{123 }$

$A_{2} = 0.0246 = 0.025 m^{2}$

Therefore, the area of second coil is ≅ 0.025 $m^{2}$

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4 years ago

What skills are used to move the body from one place to another? *

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Your muscles, bones, heart and other things in the body.

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3 years ago

Read 2 more answers

A barge floating in fresh water (rho = 1000 kg/m^3) is shaped like a hollow rectangular prism with base area A = 550 m^2 and hei

patriot [66]

Answer:

Explanation:

A )

When empty , H₀ length of barge is inside water .

volume of barge inside water = A x H₀

Weight of displaced water = AH₀ x ρ x g

Buoyant force = weight of displaced water = AH₀ ρg

B)

It should balance the weight of barge

Weight = buoyant force

Weight = AH₀ ρg

mass of barge = weight / g

weight / g = AH₀ ρ

= 550 x .55 x 1000

= 302500 kg

6 0

3 years ago