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sergeinik [125]
3 years ago
14

An object is 15 cm in front of a diverging lens with a

Physics
1 answer:
Rainbow [258]3 years ago
4 0

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

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Answer:

Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.

Maximum initial speed is 40.6 m/s

Minimum initial speed is 16.6 m/s

Explanation:

Assume this is a NET impulse so we can ignore friction.

An impulse results in a change of momentum

The impulse applied was

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p = mΔv

Δv = 8400 / 700 = 12 m/s

If the impulse was applied in the direction the car was already moving, the initial velocity was

vi = 28.6 - 12 = 16.6 m/s

if the impulse was applied in the direction opposite of the original velocity, the initial velocity was

vi = 28.6 + 12 = 40.6 m/s

Other angles of Net force would result in various initial velocities.

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