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jekas [21]
3 years ago
14

A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between

the Earth and the satellite, yet no work is done on the satellite. How is this possible?
a. No work is done if there is no contact between objects.
b. No work is done because there is no gravity in space.
c. No work is done if the direction of motion is perpendicular to the force.
d. No work is done if objects move in a circle.
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

C. No work is done if the direction of motion is perpendicular to the force.

Explanation:

We know that work is the dot product of force and displacement.

Let's take the angle between the force and the displacement = θ

W= F . d cosθ

F=Force  , d=Displacement  

If θ = 0° then W= F.d

If θ = 90° then W= 0

So we can say that when force is perpendicular to the displacement then the work done by force will be zero.

Therefore the answer is C.

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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
write a motivational speech to someone who wants to become a scientist please write this as it is for kids :) PLEASEE HELP!!!!
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Answer:

A motivational speech for a to be scientist. Crazy experiments and more you have the right to be whatever you adore. Your young yet strong everything will come along. I believe in you and so should you since your becoming a scientist you’ll acquire it soon. You’ll discover new things and sight to see when you become a scientist make sure to call me. :)

Explanation:

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Define one standard meter, one standard kilogram and one second
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6 0
3 years ago
Una persona cierra una puerta de 1 metro de ancho aplicando una fuerza de 40 [N], perpendicular a ella, a 90 [cm] de su eje de r
Damm [24]

Answer:

El módulo del torque aplicado es 36 Nm

Explanation:

En los movimientos rotatorios, la cantidad de fuerza aplicada no depende de la acción gravitacional sino del momento inercial, que es el equivalente angular de la inercia (masa) y representa la resistencia que un objeto ofrece al rotar alrededor de su eje. Cuando un cuerpo rígido rota alrededor de su eje debe considerarse , además de la masa, el radio de giro ya que estos dos factores determinan la resistencia del cuerpo a los cambios de movimiento rotatorio a través de un eje determinado.

De esta manera, se llama torque o momento de una fuerza a la capacidad de dicha fuerza para producir un giro o rotación alrededor de un punto.

En muchas ocasiones el punto de aplicación de la fuerza no coincide con el punto de aplicación en el cuerpo. En este caso la fuerza actúa sobre el objeto y su estructura a cierta distancia, mediante un  elemento que traslada esa acción de esta fuerza hasta el objeto. Entonces, el momento de una fuerza  es, matemáticamente,  igual al producto de la intensidad de la fuerza (módulo) por la distancia desde el punto de aplicación de la fuerza hasta el eje de giro:

M=F*d*sen θ

donde F es la fuerza en Newton (N), d la distancia en metros (m), θ el ángulo que forma la fuerza con el objeto al cual se le aplica la fuerza y M el momento, que se mide en Newton por metro (Nm).

En este caso:

  • F= 40 N
  • d= 90 cm= 0.9 m (siendo 100 cm= 1 m)
  • θ= 90° ya que la fuerza se aplica de forma perpendicular. Entonces sen θ= sen 90= 1

Reemplazando:

M=40 N*0.9 m* 1

Resolviendo:

M= 36 Nm

<u><em>El módulo del torque aplicado es 36 Nm</em></u>

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2 years ago
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