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miss Akunina [59]
3 years ago
8

How does the mass of an object affect its acceleration?

Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

as the mass of an object is increased, the acceleration of the object is decreased & vice-versa.

Explanation:

hope this helps!

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Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​
Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

8 0
3 years ago
I just need x isolated
mamaluj [8]

Answer:

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

Explanation:

rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0

Solve using the quadratic formula.

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

7 0
3 years ago
Read 2 more answers
What do you need to make 100 minutes charges neutral
Maru [420]
Neutrons are neutral<span> and </span>do<span> not </span>have<span> any </span>charge<span> at all. Protons carry a positive </span>charge<span>, and electrons carry the negative </span><span>charge.</span>
4 0
3 years ago
Rahul has 2 bulbs connected across two cells in a simple circuits shown. How can he make the bulbs glow dimmer?
Ray Of Light [21]

Answer:

in the parallel connection the light bulbs shine less than in the series connection

Explanation:

In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is

           P = i² R

where R is the resistance of each bulb and i the current of the circuit.

If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,

            i = i₁ + i₂

if the two light bulbs are the same

           i = 2 i₁

           i₁ = i / 2

so the power of each bulb is is

           P = i₁² R

           P = R i² / 4

           P = ¼ P_initial

Therefore we see that in the parallel connection the light bulbs shine less than in the series connection

3 0
3 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
8_murik_8 [283]

Answer:

The gplanet is 0.193 m/s^2

Explanation:

The speed of the pulse is:

v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s

v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}

where

m=mass of the wire=4 g= 4x10^-3 kg

M=mass of the object= 3 kg

Replacing values:

g=\frac{4x10^{-3}*15.24^{2}  }{3*1.6} =0.193 m/s^{2}

7 0
4 years ago
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