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3 years ago

What is true when an object is moved farther from a plane mirror?

Physics

1 answer:

musickatia [10]3 years ago

8 0

Answer:

For a plane mirror, the image distance equals the object distance, so the image distance will increase as the object distance increases

The height of the image stays the same and the image distance increases.)

Explanation:

For plane mirrors, the object distance (is equal to the image distance. That is the image is the same distance behind the mirror as the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror, you must look at a location 2 meters behind the mirror in order to view your image

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A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on th

Serhud [2]

The initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

<h3>Given:</h3>

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

∴ P = mv

P = 200 x 1

P = 200 kgm/s

∴ The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

∴ P = mv

P = 100 x 8

P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

Learn more about momentum here:

brainly.com/question/25849204

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2 years ago

Which equation describes the cosines function for right triangle?

Bogdan [553]

Answer:a

Explanation:

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3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

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3 years ago

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Butoxors [25]

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3 years ago

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juin [17]

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