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Tanya [424]
3 years ago
5

Turning the barrel of a 50-mm-focal-length lens on a manual-focus camera moves the lens closer to or farther from the sensor to

focus on objects at different distances. The lens has a stated range of focus from 0.70 m infinity.
How far does the lens move between these two extremes?
Physics
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

Explanation:

To focus object at .7m , the image distance can be measured as follows

object distance u = .7m

focal length f = .05 m

image distance v = ?

from lens formula

\frac{1}{v} -\frac{1}{u} = \frac{1}{f}

\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}

\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}

v = .054 m

= 54 mm

when the object is at infinity , image is formed at focus ie at distance of

50 mm .

So lens position from sensor  where image is formed , varies from 54 mm to 50 mm .

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Explain a situation in which you can accelerate even though your speed doesn’t change.
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Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons
yarga [219]

Answer:

q\approx 6.6\cdot 10^{13}~electrons

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

\displaystyle F=k\frac{q^2}{d^2}

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

\displaystyle q=\sqrt{\frac{F}{k}}\cdot d

Substituting values:

\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1

q = 1.05\cdot 10^{-5}~c

This charge corresponds to a number of electrons given by the elementary charge of the electron:

q_e=1.6 \cdot 10^{-19}~c

Thus, the charge of any of the spheres is:

\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}

\mathbf{q\approx 6.6\cdot 10^{13}~electrons}

5 0
3 years ago
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