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1 year ago

Derive that efficiency is the ratio of output and input work.

Physics

2 answers:

Stels [109]1 year ago

8 0

Answer:

The efficiency of a simple machine is defined as the ratio of useful work done by a machine (output work) to the total work put into the machine (input work). For ideal or perfect machine, work output is equal to the work input.

Explanation:

Hope it Helps!!!

spayn [35]1 year ago

4 0

Answer:

efficiency of a simple machine is defined as the ratio of useful work done by a machine (output work) to the total work put into the machine (input work). For ideal or perfect machine, work output is equal to the work input.

Explanation:

hope it will help you

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A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

Lina20 [59]

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop $A=\pi r^2$

$A=3.14\times 0.045^2=0.00635m^2$

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field $dB=250-350=100mT$

Emf induced is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V$

Magnitude of induced emf is equal to 0.00635 V

7 0

3 years ago

compare your previous result to the present hrf result are there any changes if yes explain your answer

garri49 [273]

Answer:

I don't know what are you saying but I don't have any results

Explanation:

$kai6417$

4 0

2 years ago

A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

1. What is the potential energy of a 5.0-kg

babymother [125]

Answer:

C. 98 J

Explanation:

The appropriate formula is ...

PE = mgh . . . . . m is mass; below, m is meters

PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2

PE = 98 J

5 0

3 years ago

State examples of a transverse wave.

laiz [17]

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

6 0

2 years ago

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Derive that efficiency is the ratio of output and input work.​

Derive that efficiency is the ratio of output and input work.