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MatroZZZ [7]
3 years ago
12

An object is placed a distance of twice the focal length away from a diverging lens. What is the magnification of the image?

Physics
1 answer:
Bas_tet [7]3 years ago
8 0

Answer:

1/3

Explanation:

We can solve the problem by using the lens equation:

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

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

Here we have a divering lens, so the focal length must be taken as negative (-f). Moreover, we know that the object is placed at a distance of twice the focal length, so

p=2f

So we can find q from the equation:

\frac{1}{q}=\frac{1}{(-f)}-\frac{1}{p}=-\frac{1}{f}-\frac{1}{2f}=-\frac{3}{2f}\\q=-\frac{2}{3}f

Now we can find the magnification of the image, given by:

M=-\frac{q}{p}=-\frac{-\frac{2}{3}f}{2f}=\frac{1}{3}

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1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

F = 15 N is the magnitude of the force

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Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

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However, in this case the displacement is

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Therefore, the magnitude of the force in this case is

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3 0
2 years ago
Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k
Yakvenalex [24]

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=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

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We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit

speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

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If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

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Then from the expression above

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E = 6.90 x 10^-5 x 12388.32 V

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Rudik [331]

Answer:

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3 years ago
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Explanation:

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