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Anastasy [175]
2 years ago
13

Which type of force does a window washer want acting on him when trying to keep the same position? Question 5 options: large sma

ll balanced unbalanced.
Physics
1 answer:
taurus [48]2 years ago
8 0

The balanced force is required to be acting on the window washer to keep it in the same position.

<h3>What is Force?</h3>

The external effort causing the change in the state of motion of any object is known as force.

Forces are generally of two types namely balanced force and unbalanced force.

When an object is applied with more than one force, such that net force on the object is zero and due to this the state of motion of object remains unaffected, then such type of force is called balanced force.

Similarly, the forces on the object having some net magnitude and causing some motion to the object are known as unbalanced forces.

So as per the given problem, for keeping the window washer in the same position, it is required to apply with a balanced force.

Thus, we can conclude that balanced force is required to be acting on the window washer to keep it in the same position.

Learn more about the types of forces here:

brainly.com/question/13774406

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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
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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

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Let;

Image distance, i,=?

Image height, h'=?

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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}

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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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