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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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Which of the following most accurately represents John Dalton’s model of the atom? A. a tiny, solid sphere with an unpredictable
aleksley [76]
A and c are the answersss
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2 years ago
Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli
Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: sin \theta=\frac{m\lambda}{\alpha}

Given m=3,\lambda=6.33\times 10^-^7 and \alpha=0.00015:

#Substitute our variable values in the minima equation to obtain \theta:

\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad

#draw a triangle to find the relationship between \theta, y \ and L.

tan(\theta)=y/L               #where y=4.05cm

L=y/tan(\theta)=3.20

Hence the screen is 3.20m from the split.

b.  To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

sin \theta=\frac{m\lambda}{\alpha}, where m=4 given that it's the minima with the smallest distance.

sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad

#we then use tan(\theta)=y/L to calculate L=4.5cm

Then from the equation subtract y_3 from y:

4.50cm-4.05cm=0.45cm

Hence, the distance \bigtriangleup y is 0.45cm

8 0
2 years ago
Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −
guapka [62]

Answer:

Explanation:

Given two vectors as follows

E₁ = 13.5 i -12 j

E₂ = -7.4 i - 4.7 j

Resultant E = E₁ + E₂

= 13.5 i -12 j -7.4 i - 4.7 j

E = 6.1 i - 16.7 j

a ) X component of resultant = 6.1 N

b ) y component of resultant = -16.7 N

Magnitude of resultant = √ ( 6.1² + 16.7² )

= 17.75 N

d ) If θ be the required angle

tanθ = 16.7 / 6.1 = 2.73

θ = 70° .

counterclockwise = 360 - 70 = 290°

6 0
2 years ago
A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the
cupoosta [38]

Answer: deceleration of 1.904\ rad/s^2

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. a=\alpha \cdot r

Using the equation of motion

v^2-u^2=2as\\

Insert v=0,u=20,s=300

0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2

Write acceleration as a=\alpha \cdot r

-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2

So, the car must be experiencing the deceleration of 1.904\ rad/s^2.

4 0
2 years ago
You are provided with three polarizers with filters making angles of (A) 90 ​∘ ​​ , (B) 180 ​∘ ​​ and (C) −45 ​∘ ​​ with respect
irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) 90^{\circ}:

The light intensity after passing through the first polarizer is I_{o} and this intensity does not depend on the angle of the polarizer.

Consider 90^{\circ} with the vertical, the intensity is given by:

I = I_{o}cos^{2}90^{\circ}

I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}

(B) 180^{\circ}:

Suppose the second polarizer is  45^{\circ} with the vertical.

Now, intensity through the second polarizer:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)

I' =  \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}

Now, if we consider the second polarizer to be 180^{\circ},

I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0

(C) - 45^{\circ}:

Now,

Intensity through the third polarizer, if it is 180^{\circ} with the vertical:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))

I' = \frac{I_{o}}{8}

5 0
2 years ago
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