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tamaranim1 [39]
3 years ago
15

Camera lenses are described in terms of

Physics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

The position of the image from the camera lens is approximately 121.1 mm

Explanation:

The given parameters of the lens are;

The specification of the camera lens = 120 mm

Therefore, the focal length of the camera lens, f = 120 mm = 0.12 m

The distance of the object from the camera, d_o = 13 m

The lens equation for finding the position of the image is given as follows;

\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}

Where;

d_i = The position of the image from the camera lens

Therefore, by plugging in the known values, we have;

\dfrac{1}{13} = \dfrac{1}{0.12} + \dfrac{1}{d_i}

\dfrac{1}{d_i}  = \dfrac{1}{0.12} -  \dfrac{1}{13} = \dfrac{13 - 0.12}{0.12 \times 13} = \dfrac{12.88}{1.56} = \dfrac{322}{39}

\therefore d_i = \dfrac{39}{322}  \approx  0.1211

The position of the image from the camera lens, d_i = 0.1211 m = 121.1 mm

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Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

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then

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Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

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h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m

3 0
3 years ago
Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni
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Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T

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Answer:

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