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icang [17]
3 years ago
10

Gravitational force is directly related to mass and inversely related to distance. So the bigger the masses the greater the forc

e and the greater the distance the smaller the force. According to the model, which sphere has the greatest gravitational attraction to sphere D?
Q:
A:
Physics
1 answer:
Iteru [2.4K]3 years ago
3 0

Answer:Well according to the formula :

Fg = G • m1 • m2

__________

r^2

If one were to increase the masses of the objects it will have a direct impact on the gravitational force as well, as it increase the gravitational force. It is directly proportional to mass.

Distance on the other hand is inversely proportional to the gravitational force, basically increasing the distance between two objects of substantial mass causes a decrease of attraction or gravitational force between those 2 bodies of mass.

Explanation:

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A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0055 m/s2 less than that at sea leve
pshichka [43]
Gs*rs^2 = gm*rm^2 
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
7 0
4 years ago
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

a'=4\times \dfrac{v^2}{R}

a'=4\times a

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(1
tekilochka [14]

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>m</em> <em>a</em>

∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N

Let <em>f</em> be the unknown force. Then

∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>

=>   <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N

8 0
3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

3 0
3 years ago
What is the stretch when you pull with a force of 25 N on a spring with a spring constant of 8 N/m? *
Pani-rosa [81]

Hooke's Law

\tt F=k.\Delta x

k = spring constant

x = stretch

F = force

Input the value

\tt \Delta x=\dfrac{F}{k}=\dfrac{25}{8}=3.125\rightarrow 3.13\:m

7 0
2 years ago
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