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kiruha [24]
1 year 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]1 year 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]1 year 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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A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
1 year ago
1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and
saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

   Centripetal acceleration  = \frac{v^{2} }{r}

   v is the velocity of the body

    r is the radius of the circle

  putting in the parameters:

   Centripetal acceleration = \frac{4^{2} }{0.75}

    Centripetal acceleration = 10.67m/s²

Centripetal force = m  \frac{v^{2} }{r}

          m is the mass

 Centripetal force = mass x centripetal acceleration

                              = 3 x 10.67

                              = 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

4 0
1 year ago
The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy
Lisa [10]
Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J
3 0
2 years ago
Read 2 more answers
A 60 kg student in a rowboat on a still lake decides to dive off the back of the boat. The studen'ts horizontal aceleration is 2
TiliK225 [7]
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.

So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.

Force on the student = 60 kg * 2.0 m/s^2 = 120 N.

=> horizontal force exerted by the student on the boat = 120 N

Answer: option d. 120 N. toward the back of the boat.

Of course it is toward the back because that is where the student jumped from..
4 0
2 years ago
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