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3 years ago

A transformer is used to convert mechanical energy to electric energy

Physics

1 answer:

PtichkaEL [24]3 years ago

3 0

No it isn't. That's a description of a generator.

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Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.

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No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.

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3 years ago

What does g-force mean?

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3 years ago

Whats the measure of how much energy a sound wave carries, the loudness of a sound?

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3 years ago

Can someone please answer?

nevsk [136]

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2 years ago

Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica

Aleks04 [339]

We have that the magnification of each focal length is given respectively as

A) has $u=3\frac{f}{2}$

B) has $u=4\frac{f}{3}$

C) has $u=5\frac{f}{4}$

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

$M=\frac{-v}{u}$

Therefore

$v=2u$

For A

$M=-2$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}$

Therefore

$u=3\frac{f}{2}$

For B

$M=-3$

Therefore

$v=3u$

Where

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}$

Therefore

$u=4\frac{f}{3}$

For C

$M=-4$

Therefore

$v=4u$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}$

Therefore

$u=5\frac{f}{4}$

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has $u=3\frac{f}{2}$

B has $u=4\frac{f}{3}$

C has $u=5\frac{f}{4}$

For more information on this visit

brainly.com/question/14468351

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2 years ago