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2 years ago

Calculate The pressure produced by a force of 392 N acting on an area of 8.0 m^2

Physics

1 answer:

Nonamiya [84]2 years ago

5 0

Answer:

Explanation:

500-0.05=499.95

500-100+5=405

100+100=10000

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A lab cart is loaded with different masses and moved at various velocities

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A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

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3 years ago

The spherical annulus of cometary nuclei surrounding the solar system is called the ___.

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Answer:

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Explanation:

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3 years ago

Which part of the drum vibrates when it produces sound?

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3 years ago

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3 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

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3 years ago