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Vedmedyk [2.9K]
4 years ago
6

A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc

e opposing the motion?
Physics
1 answer:
docker41 [41]4 years ago
3 0

The frictional force is 15 N

Explanation:

In order to solve this problem, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we are just interested in the situation along the horizontal direction. The net force acting on the block is:

\sum F = F_a -F_f

where

F_a = 40 N is the applied force (forward)

F_f is the frictional force (backward)

m = 5 kg is the mass of the block

a=5 m/s^2 is the acceleration

Substituting into the equation and solving, we find the frictional force:

F_a-F_f = ma\\F_f = F_a-ma=40-(5)(5)=15 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

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a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

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

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

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

where:

y' is the height of the image

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q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

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(3.25 x 10^5) + (7.5 x 10^4) = 400000

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