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Deffense [45]
3 years ago
14

HELP PLEASE!!! 30+ points!!

Physics
1 answer:
GarryVolchara [31]3 years ago
4 0

1) 9.26 cm

Explanation:

The focal length of a plane mirror is virtually infinite. Considering the lens equation,

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

where f is the focal length, p is the object distance, q the image distance. If we replace f with infinity, we get

q=-p

The magnification equation states that

y' = -\frac{q}{p}y

where y is the size of the object and y' the size of the image. Substituting q=-p, we get

y'=y

this means that the image produced by a plane mirror is always:

- Upright (y' is positive)

- The same size as the object

In this case, we have a book of height 9.26 cm (y=9.26 cm). This means that the magnitude of the size of the image (y') will be 9.26 cm as well.

2) 22.7 cm

As we said before, due to the infinite focal length of a plane mirror,

q=-p

this means that the image produced by a plane mirror is always:

- Virtual (because q is negative)

- At the same distance from the mirror as the object

In this case, we have a book placed at 22.7 cm from the mirror (p=22.7 cm). This means that the magnitude of the distance of the image from the mirror (q) will be 22.7 cm as well.

3) 1.60 m/s

We said previously that the image produced by a plane mirror is always at the same distance from the mirror as the real object. This implies that whenever we move the object toward/away from the mirror, the distance p will alway remain equal to the distance q. But this also means that the object and the distance are moving toward/away from the mirror at the same speed.

Therefore, since in this case the person is moving away from the mirror at 1.60 m/s, the image will also move away at a speed of 1.60 m/s.

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3 years ago
14. A ball initially at rest rolls without slipping
elena-s [515]

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v = √(10gh/7)

Explanation:

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3 years ago
A Boeing 737—a small, short-range jet with a mass of 51,000 kg— sits at rest at the start of a runway. The pilot turns the pair
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Answer:

67000N

Explanation:

We solve for the acceleration using the the 3rd constant-acceleration equation.

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We have the displacement to be

∆x = Xf - Xi = 940m

Vx = 70m/s

The acceleration = (70m/s)²/2(940m)

= 4900/1880

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51000kg x 2.61m/s²

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