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Tom [10]
3 years ago
14

If gravity between the sun and Earth suddenly vanished; describe the expected Earth motion.

Physics
1 answer:
Daniel [21]3 years ago
6 0

Answer:

When the gravity between the Sun and the Earth suddenly vanishes, then Earth will keep moving in a straight line in a direction where it was moving at the moment when gravity vanished.

Explanation:

The gravitational force between Sun and Earth is given by

F = G\frac{M_{1}M_{2} }{R^{2}}

Where F is the gravitational force between the Sun and the Earth. M₁ and M₂ are the masses of Sun and Earth and R is the distance between them.

If we assume that the gravity between the sun and Earth suddenly vanishes, then there would not be any force between the Sun and the Earth and the Earth will keep on moving in a straight line. This is endorsed by the Newton's first law that a body in motion remains in motion if no external force is acting on it.

The direction of Earth's motion will be determined by the previous direction of the motion that is the moment when the gravity was vanished.

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What did Edwin Hubble discover about galaxies that allowed him to determine the expansion rate of the universe?
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3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires
svlad2 [7]

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

Y=\frac{F \times L}{A\times \delta L}

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

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3 years ago
A single point charge is placed at the center of an imaginary cube that has 19 cm long edges. The electric flux out of one of th
OverLord2011 [107]

Answer:

The amount of charge at the center is -132.7 nC

Solution:

As per the question:

Electric flux from one face of cube, \phi_{E} = - 2.5 kN.m^{2}/C

Edge, a = 19cm

Since, a cube has six faces, thus total flux from all the 6 faces = 6\times (-2.5) = - 15 kN.m^{2}/C

Also, from Gauss' law:

\phi_{E,net} = \frac{1}{\epsilon_{o}}Q_{enc}

Q_{enc} = 8.85\times 10^{- 12}\times - 15\times 10^{3}

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3 years ago
A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of
Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

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