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exis [7]
3 years ago
6

A box sits on a table. A long arrow labeled F subscript P = 22 N points right. A short arrow labeled F subscript f = ? N points

left. A man pushes on the box from the left. The net force acting on the box is 18 N. What is the force of friction?
–22 N
–4 N
4 N
22 N

Physics
2 answers:
anygoal [31]3 years ago
8 0

Answer:

b -4 N

may the force be with you

dexar [7]3 years ago
7 0

Answer:

b-4 N

Explanation:

I think thats right

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A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant
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In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

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3 years ago
Theweight ofof body is 420N .Calculate its mass​
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42.87

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<u>OR</u>

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420/9.8 which equals 42.87.

6 0
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A gas always spreads out to fill all available space.<br> True or False <br> Science Hw
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true

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6 0
3 years ago
Read 2 more answers
An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c
Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

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

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0
3 years ago
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