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k0ka [10]
3 years ago
11

True or false gravitational pull decreases with an increase of distance between two objects​

Physics
1 answer:
ohaa [14]3 years ago
5 0

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

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a 150 N force is used to pull a wooden box across a wooden surface at a constant velocity. what is the mass of the box?
IRINA_888 [86]

Answer:

The mass of the box:

m =  60 kg

Explanation:

Given:

F = 150 N

g = 10 m/s²

_________

m - ?

Coefficient of friction wood on wood:

μ = 0.25

Friction force:

F₁ = μ*m*g

Newton's Third Law:

F = F₁

F = μ*m*g

The mass of the box:

m = F / ( μ*g) = 150 / (0.25*10) =  60 kg

7 0
1 year ago
The mass of an atom is
suter [353]

Answer:

Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.

I hope this is the answer you were looking for :D

4 0
2 years ago
Read 2 more answers
a light ray traveling in air enters a second medium and its speed slows to 1.71 × 10^8 meters per second. what is the absolute i
givi [52]
The absolute refractive index is equal to the speed of light of the wave in air divided by the speed of light in the second medium. This means that it is equal to 3 x10^8 / 1.71 x10^8. This means the answer is 1.75
8 0
3 years ago
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The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov
Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

4 0
3 years ago
Name the apparatus used to measure friction force? PLEASE HELP AS SOON AS POSSIBLE
melisa1 [442]
There is no apparatus for it. It is either use something like ruler and table and rub together or rub our hands, and friction force will be showed.
8 0
3 years ago
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