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3 years ago

Is a force pushing up on an object as gravity is pulling down?

Physics

1 answer:

aliya0001 [1]3 years ago

6 0

Answer:

Logically yes, because Newton's Third law state "When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."

If force wasn't pushing up then neither gravity is pulling down.

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A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air ha

Tamiku [17]

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

θ = sin⁻¹ n₂ / n₁

let's calculate

θ = sin⁻¹ (1.00 / 1.50)

θ = 41.8º

4 0

3 years ago

Which of the following solid fuels has the highest heating value?

Makovka662 [10]

I wanna say its A . I could be wrong but im almost 100 percent sure that its A wood

4 0

3 years ago

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the book is

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied on the book, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the book is

F=44N

For more information on this visit

brainly.com/question/23379286

8 0

2 years ago

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di

vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi (\frac{D}{2})^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= \frac{d}{\pi}$

$r_e = \sqrt{\frac{d}{\pi} }$

replacing d = $\frac{\pi (\frac{D}{2})^2 }{10, 000}$ in the equation above; we have:

$r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }$

$r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}$

$r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

4 0

3 years ago

The message refers to which of the following?

Oliga [24]

The message is the information being communicated from one place to another.

It used to be called the "intelligence". But as time went on, it became
harder to ignore the obvious fact that that was going too far, and the
label was changed to the more IQ-neutral "message".

7 0

3 years ago