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Sidana [21]
3 years ago
7

Which factors affect the gravitational force between objects? Check all that apply.

Physics
2 answers:
storchak [24]3 years ago
6 0

Answer:

the answer is C and D

Explanation:

did it on edge

hope u have better day than me

hope this helps

drek231 [11]3 years ago
4 0

Explanation:

distance between both objects and masses of the objects

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Mention one life application on density ​
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One well-known application of density is determining whether or not an object will float on water. If the object's density is less than the density of water, it will float; if its density is less than that of water, it will sink.In fact, submarines dive below the surface of the water by emptying their ballast tanks

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2 years ago
What are the frequencies of EM spectrum
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The EM spectrum has no limits.  Any frequency you can imagine
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3 years ago
Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electr
Eduardwww [97]

Answer:

D=1693742.7m

Explanation:

For sound waves we have v=d/t where v is the speed of sound and d the distance between the astronauts, while for electromagnetic waves we have c=D/t where c is the speed of light and D the distance between the spaceship and Earth. <em>We have written both times as the same</em> because is what is imposed by the problem, so we have t=d/v=D/c, which means:

D=\frac{dc}{v}

And for our values:

D=\frac{(2m)(299792458m/s)}{354m/s}=1693742.7m

5 0
3 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
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Why are lamps connected in parrells rather than in a series in domestic circuit
Lilit [14]

Answer:

so that each component has the same voltage.

Explanation:

5 0
3 years ago
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