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kiruha [24]
2 years ago
5

Consider a solid, rigid spherical shell with a thickness of 100 m and a density of 3900 kg/m3 . the sphere is centered around th

e sun so that its inner surface is at a distance of 1.50×1011 m from the center of the sun. what is the net force that the sun would exert on such a dyson sphere were it to get displaced off-center by some small amount?
Physics
2 answers:
Ne4ueva [31]2 years ago
5 0
0 N

According to sources, the most probable answer to this query is zero (0) newtons. Since there is no more acting force on the objects and it reaches net force, obviously the answer would be zero. 

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
m_a_m_a [10]2 years ago
4 0

Answer:

Here in the given situation that Sun is surrounded by shell of given thickness so that the force on the Sun due to shell must be zero and hence there is no displacement on the center of the shell

Explanation:

As per the concept of Shell Theory we know that when a mass is placed inside the hollow shell then the force of attraction between shell and the given object is always zero

It is because the force of gravitation on the object is due to vector sum of the force due to each part of the shell. So when we vectorially add all the components of the force due to shell then in that case the net force on the object will be zero

Here it is the same situation that Sun is surrounded by shell of given thickness so that the force on the Sun due to shell must be zero and hence there is no displacement on the center of the shell

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When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as
Dafna11 [192]
When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>
3 0
2 years ago
a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w
AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, M=75\ kg

Mass of the ball is, m=4\ kg

Initial velocity of the man is, u_m=0\ m/s(rest)

Initial velocity of the ball is, u_b=0\ m/s(rest)

Final velocity of the ball is, v_b=3.50\ m/s

Final velocity of the man is, v_m=?\ m/s

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of man and ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Now, initial momentum = final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0
2 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
Ne4ueva [31]
A)  f = 1.8 rev/s = 2 Hz 
<span>T = 1 / f = 0.55s

B)  not really sure..srry

C)  </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
</span>
Hope this helps a little at least.. :)

5 0
2 years ago
A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor
Alex17521 [72]

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0
2 years ago
A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy
skad [1K]

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

\theta= \omega t

where

\omega is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of \omega is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

(b) The girl has greater linear speed.

Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the point from the centre of the merry-go-round

In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of \omega is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.

3 0
2 years ago
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