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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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What is NOT one of the three primary resources that families have to reach financial goals?
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What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
8 0
2 years ago
Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el
Rina8888 [55]
<span>this  may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
8 0
2 years ago
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About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot
9966 [12]

Answer:

2.44156\times 10^{13}\ m^3

29010.53917 m

Explanation:

\rho = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = \dfrac{d}{2}=\dfrac{10}{2}=5\ km

v = Velocity = 11 km/s

H_v = Heat vaporization of water = 2.26\times 10^6\ J/kg

\Delta T = Change in temperature = 100-20

Mass is given by

m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg

The kinetic energy is

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J

Heat is given by

Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg

Mass of water is 2.44156\times 10^{16}\ kg

Volume is \dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3

Amount of water is 2.44156\times 10^{13}\ m^3

If it were a cube

h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m

The height of the water would be 29010.53917 m

4 0
2 years ago
At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

8 0
2 years ago
A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w
bezimeni [28]

Answer:

The ball was in air for 3.896 s

Explanation:

given,

g = 9.8 m/s², acceleration due to gravity,

If the launch angle is 45°, the horizontal range will be maximum.

The horizontal and vertical launch velocities are equal, and each is equal to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The time to attain maximum height is one half of the time of flight.

v = u + at                     ∵ v = 0 (max. height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The time of flight is twice of the maximum height time

2 t₁ = 3.896 s

The horizontal distance traveled is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in air for 3.896 s

8 0
2 years ago
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