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Over [174]
1 year ago
9

If Earth were 10.0 times farther away from the Sun than it is now, how many times weaker would the gravitational force between t

he Sun and Earth be
Physics
1 answer:
choli [55]1 year ago
5 0

If Earth were 10.0 times farther away from the Sun than it is now, 100 times weaker would the gravitational force between the Sun and Earth.

What is Gravitational Force?

According to Newton's universal law of gravitation, The force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

What causes gravitational force?

Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That's what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here.

Learn more about gravitational force:

brainly.com/question/862529

#SPJ4

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A fishbowl has a circular opening with a diameter of 13 cm. The fishbowl sits upright on a table in a magnetic field of 0.00110
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Answer:

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Explanation:

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How do you calculate energy lost due to friction in an experiment?
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A treadmill get it? but its   Ff * d cos theta

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3 years ago
A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte
zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

5 0
3 years ago
Ned tightens a bolt in his car engine by exerting 15 N of force on his wrench at a distance of 0.45m from the fulcrum. How much
____ [38]

Answer:

T = 6.75\,N\cdot m

Explanation:

Vectorially speaking, torque is the cross product between force and distance from fulcrum. Its magnitude is equal to the following expression:

T = F\cdot r \cdot \sin \theta

T = F_{\bot} \cdot r

Let assume that force is perpendicular to the distance from the fulcrum. So, the torque needed to turn the bolt is:

T = (15\,N)\cdot (0.45\,m)

T = 6.75\,N\cdot m

8 0
3 years ago
For elliptical obits: the direction of the velocity of the satellite is _______________________ (always, seldom, never) perpendi
alexandr1967 [171]

Answer:

For elliptical orbits: seldom

For circular orbits: always

Explanation:

We start by analzying a circular orbit.

For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.

Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.

So for a circular orbit,

the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.

Now we analyze an elliptical orbit.

An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.

Therefore, for an elliptical orbit,

the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.

7 0
3 years ago
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