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Shalnov [3]
1 year ago
9

The focal length of the lens of a simple camera is 40.0 mm. how far must the lens be moved to change the focus of the camera fro

m a person 25 m away to one that is 4.0 m away?
Physics
1 answer:
Minchanka [31]1 year ago
4 0

.05 cm far must the lens be moved to change the focus of the camera from a person 25 m away to one that is 4.0 m away

The focal length of a thin lens in air is the separation between its major foci, also known as focal points, and its center. The focal length is the distance at which a beam of collimated light will be concentrated to a single spot for a converging lens (for instance, a convex lens).

If v is the image distance in the first scenario

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

The second instance

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

Therefore, the lens must be advanced by 5.06 - 5.01 =.05 cm.

Learn more about  lens here brainly.com/question/17086993

#SPJ4.

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<em>Hope this helps :)</em>

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a flowerpot falls from a windowsill 25m above the sidewalk how fast is the flowerpot moving when it strikes the ground
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s = 25m

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3 0
3 years ago
A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma
goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

F=ma

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

F=ma

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

x(t)=\frac{1}{2}a_g t^2

where a_g = 0.098 m/s^2 is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

x'(t)=15-\frac{1}{2}a_s t^2

where a_s=0.8 m/s^2 is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s

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3 years ago
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