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3 years ago

Does the moon actually change shape as it revolves/rotates around the Earth? Explain your answer

Physics

1 answer:

rusak2 [61]3 years ago

3 0

Answer:

The moon doesn’t change shape on its own.

Explanation:

Shapes of moon that we observe is based on the different perspectives of view from the earth and position of moon with respect to the sun. The changes arise due to the rotation of earth on its own axis as well as the revolution of moon on its orbit. The moon doesn’t have any light of its own.

It just reflects off the light from the sun. Due to tidal locking phenomenon one face of the moon permanently faces the sun. Because of the changes in position of moon with respect to the sun the moon is lighted up variably giving rise to various phases like new moon, full moon, crescent etc.

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Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one

kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ = - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0

3 years ago

Which is true of electricity generated both from coal and from nuclear reactions?

aleksklad [387]

They both release greenhouse gases, I think

8 0

3 years ago

Read 2 more answers

A particle with charge q, mass m, and initial speed v0 in the x direction enters a region where the electric field is uniform in

Nezavi [6.7K]

Answer:

S = 1/2 Vo t + 1/2 a t^2 = d time for particle to travel distance d

F = E q force acting on particle

a = F / m = E q / m

d = Vo t + E q / (2 m) t^2

One would need to solve the quadratic equation shown to find the time t

t^2 + (2 m) / E q * V0 t - (2 m) / E q * d = 0

or t^2 + A V0 t - A d = 0 where A = (2 m) / E q

8 0

2 years ago

How would changing the proportions of substances in an alloy change its properties

Aloiza [94]

It would make them different. they wouldn't be the same.

6 0

3 years ago

Read 2 more answers

A bowling ball weighing 71.2 N (16.0 lb) is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and release

Elina [12.6K]

Answer:

a) For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

b) For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

Explanation:

For this case we have the following data given:

$W= 71.2 N$ represent the weigth for the object

$R=3.8 m$ the length of the rope or the radius

$v= 4.2 m/s$ represent the velocity of the bowling ball

Part a

For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

Part b

For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

7 0

2 years ago