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lora16 [44]
4 years ago
8

PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS​

Physics
1 answer:
butalik [34]4 years ago
8 0

Answer:

g=GM/R^2

Universal Gravutation Constant:

f=GM×m/R^2

Force can be also expressed as

f=m×g

so,

mg=GMxm/R^2

The m gets cancelled so

g=GM/R^2

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True or False. Recent findings lend strong support to the theory that a black hole lies at the center of the Milky Way and of ma
Genrish500 [490]

Answer:

True

Explanation:

In 2019 was taken the first image of a supermassive black hole in the center of the galaxy M87 with the Event Horizon Telescope, which is a network of radio telescope located in different points of the Earth, with the purpose of making a telescope of the size of the Earth.

A radio telescope is an antenna that is capable to perceive the light in the radio part of the electromagnetic spectrum¹.

It is important to notice that in the picture what it can be seen is the effect that the black hole has in the nearby stars.

Key terms:

¹Electromagnetic spectrum: decomposition of light in its different wavelengths (from radio waves to gamma rays).

7 0
3 years ago
What is Physics? How is Physics used in your everyday life? Give at least two examples and explain your answer. You need at leas
ikadub [295]

Answers:

The transportation industry is no stranger to the manipulation of everyday physics. Cars and trains utilize the wheel, which provides a smooth, steady motion.

The ears hear sounds which occur through the movement of air molecules, and the chemistry that drives all of biology depends on the physics of energy and molecules. Every day, for example, plants absorb sunlight, water, and carbon dioxide, creating glucose and releasing oxygen as a byproduct.

Brainlist pls!

3 0
3 years ago
A mass m neutron has elastic collision with a mass m'
hoa [83]

Answer:

The neutron loses all of its kinetic energy to nucleus.

Explanation:

Given:

Mass of neutron is 'm' and mass of nucleus is 'm'.

The type of collision is elastic collision.

In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.

Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.

Now, final kinetic energy of the system will be equal to the initial kinetic energy.

Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.

Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.

Therefore, the neutron loses all of its kinetic energy to nucleus.

5 0
3 years ago
A trapeze artist swings in simple harmonic motion with a period of 3.8 s.
pogonyaev
As we know that time period of simple pendulum is given as

T = 2π √L/g

here we know that

T = 3.8 s

now from above equation we know that

T² = 4π² (L/g)

now on rearranging the above equation we will have

L = gT² / 4π²

now plug in all data into it

L = (9.8) (3.8)² / (4) (3.14)²

so the length of the cable must be 3.6 m
7 0
3 years ago
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0
4 years ago
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