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Assoli18 [71]
3 years ago
7

If the mass of the Earth somehow increased with no change in radius, your weight would

Physics
2 answers:
Margaret [11]3 years ago
5 0

Answer:

increase also

Explanation:

The weight of a person is equal to the gravitational pull exerted by the Earth on the person:

F=G\frac{mM}{R^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M is the mass of the Earth

m is the mass of the person

R is the Earth's radius

We notice that the weight is directly proportional to the mass of the Earth. Therefore, if the mass of the Earth M increases, and the radius R does not change, the weight of the person increases as well.

Sever21 [200]3 years ago
5 0

<u>Answer</u>

Increases also.

The force due to gravity is given by,

F = GM.m/r²

Where G is a constant of proportionality

∴ F ∝ M.m/r²

When r remains constant, force due to gravity, F, will be;

F ∝ M.m

Where M is the mass of the earth and m is your mass

Since your mass does not change, we are going to have;

F ∝ M.

This means the weight F is directly proportional to the mass of the earth. when it increases the your weight also increases.

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A.stationary <br> B. Accelerating <br> C. Decelerating <br> D. Moving at constant speed
V125BC [204]

Answer:

ACCELERATING OR DECELERATING

Explanation:

I'M NOT SURE

7 0
2 years ago
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
The suspension system mounts the car's wheels<br> solid on the frame. True or false?
nadezda [96]
Suspect and mounts are a solid frame. True.
3 0
3 years ago
Which cells or organs are considered to be part of both the immune and lymphatic systems? Select all that apply.
Leviafan [203]

Answer:

lymph nodes

tonsils and adenoids

thymus

Explanation:

-Arteries are the blood vessels that take the blood that contains oxygen from the heart to the tissues and are part of the circulatory system.

-Lymph nodes are glands that take care of filtering the fluid that goes through the lympathic system and are also important for the functioning of the immune system.

-Capillaries are blood vessels that connect the veins and arteries and are part of the circulatory system.

-Tonsils and adenoids are located in the throat and they help protect the body from diseases and they are part of immune system and the lympathic system.

-Veins are the vessels that take the blood to the heart and they are part of the circulatory system.

-Thymus is an organ in which the T cells develop and they help protect the body against virus and bacteria and it is part of the immune and lympathic systems.

According to this, cells or organs that are considered to be part of both the immune and lymphatic systems are:

lymph nodes

tonsils and adenoids

thymus

8 0
2 years ago
What is the acceleration of an object with a mass of 15 kg and a coefficient of friction of 0.18
11111nata11111 [884]

Answer:

a = 1.764m/s^2

Explanation:

By Newton's second law, the net force is F = ma.

The equation for friction is F(k) = F(n) * μ.

In this case, the normal force is simply F(n) = mg due to no other external forces being specified

F(n) = mg = 15kg * 9.8 m/s^2 =  147N.

F(k) = F(n) * μ = 147N * 0.18 = 26.46N.

Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.

Thus, F(net) = F(k) = ma

26.46N = 15kg * a

a = 1.764m/s^2

7 0
2 years ago
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