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klio [65]
3 years ago
9

A 12-inch object is placed 30 inches in front of a plane mirror. A ray of light from the object strikes the mirror at a 45-degre

e angle. How tall is the image produced by the mirror?
A) 12 inches
B) 30 inches
C) 24 inches
D) 45 inches
Physics
2 answers:
JulsSmile [24]3 years ago
4 0

Answer: A. 12inches

Explanation:

One of the characteristics of an object placed in front of a plane mirror is that the object distance(u) is equal to the image distance(v) no matter the position of the object from the mirror.

Since object distance is 30 inches, the image distance will be 30inches as will i.e u = v

Using the relationship to get the height of the image

v/u = h1/h0

Where h1 is image height and h0 is object height.

h0 = 12inches

Substituting the values given in the formula we have

30/30 = h1/12

h1 = 12inches.

This means that the image produce by the object is also 12inches tall.

gtnhenbr [62]3 years ago
3 0

I believe that it is C.24 inches

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Answer:

a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)

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Explanation:

a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly

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The mass of the person riding on the Ferris wheel, the passenger  = 55 kg

Therefore, we have;

The angular speed, ω = Δθ/Δt = 2·π/(28)

From which we have;

Centripetal force, F_c = m × ω² × r

Substituting the known values, we have F_c = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N

The centripetal force, F_c = 19.94 N always acting outward from the center

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The weight of the passenger = 55 kg × 9.8 m/s² = 539 N

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At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards

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The net force, which is the apparent weight of the passenger at the top F_{NET_{Top}} = 539 N - 19.94 N ≈ 519.06 N

Apparent weight at the top ≈ 519.06 N

At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards

Therefore;

The net force at the bottom, which is the apparent weight of the passenger at the bottom F_{NET_{bottom}} = 539 N + 19.94 N ≈ 558.94 N

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