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Which is a result if using a machine

1 answer:

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There are many types of machine what type well for every day machines like treadmills is sweating, dehydration, loss of calories or fat.

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A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

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Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

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If the amplitude of the resultant wave is twice as great as the amplitude of either component wave, and the wave exhibits reinfo

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Be in phase with each other! otherWise B

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a spherical mirror produces a magnification of -1 on a screen placed at a distance of 40cm from the mirror i) write the type of

Darya [45]

Answer:

f =-20 cm

Explanation:

Given that,

The magnification of a spherical mirror, m = -1

The image distance, v = 40 cm (for negative magnification)

The magnification of a concave mirror is negative. The mirror showing -1 magnification is a concave mirror.

Let f be the focal length of the mirror. We know that,

$m=\dfrac{-v}{u}\\\\-1=\dfrac{-v}{u}\\\\v=u$

Object distance, u = -40 cm

Using mirror's formula i.e.

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\\\\f=\dfrac{1}{\dfrac{1}{-40}+\dfrac{1}{-40}}\\\\f=-20\ cm$

So, the focal length of the mirror is 20 cm.

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When we stretch a spring, which two factors affect how much elastic

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Mass and gravitational field

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