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Fudgin [204]
2 years ago
13

1. Calculate the magnitude of the gravitational force between the Earth and an m = 5.50 kg mass on the surface of the Earth. The

distance to the center of the Earth from the surface is 6.37×103 km and the mass of the Earth is 5.98×1024 kg.
2. Calculate the magnitude of the gravitational force between the Moon and an m = 5.50 kg mass on the surface of the Earth nearest to the moon. The distance to the center of the Moon from the surface of the Earth is 3.76×105 km and the mass of the Moon is 7.36×1022 kg.
3. Calculate the ratio of the magnitude of the gravitational force between an m = 5.50 kg mass on the surface of the Earth due to the Sun to that due to the Moon. The mass of the Sun is 1.99×1030 kg and the distance from the center of the Sun to the surface of the Earth is 1.50×108 km.
Physics
1 answer:
PIT_PIT [208]2 years ago
5 0

1. The magnitude of the gravitational force between the Earth and an m is  54.1 N.

2. The magnitude of the gravitational force between the Moon and an m is  1.91 x 10⁻⁴ N.

3. The ratio of the magnitude of the gravitational force between an m on the surface of the Earth due to the Sun to that due to the Moon is 169.6.

<h3>Gravitational force between Earth and mass, m</h3>

The gravitational force between Earth and mass, m is calculated as follows;

F(Earth) = Gm₁m₂/R²

F(Earth) = (6.67 x 10⁻¹¹ x 5.5 x 5.98 x 10²⁴)/(6,370,000)²

F(Earth) = 54.1 N

<h3>Gravitational force between Moon and mass, m</h3>

F(moon) = Gm₁m₂/R²

F(moon) = (6.67 x 10⁻¹¹ x 5.5 x 7.36x 10²²)/(3.76 x 10⁸)²

F(moon) = 1.91 x 10⁻⁴ N

<h3>Gravitational force between Sun and mass, m</h3>

F(sun) = Gm₁m₂/R²

F(sun) = (6.67 x 10⁻¹¹ x 5.5 x 1.99x 10³⁰)/(1.5 x 10¹¹)²

F(sun) = 0.0324 N

<h3>Ratio of F(sun) to F(moon)</h3>

= 0.0324/1.91 x 10⁻⁴

= 169.6

Thus, the magnitude of the gravitational force between the Earth and an m is  54.1 N.

The magnitude of the gravitational force between the Moon and an m is  1.91 x 10⁻⁴ N.

The ratio of the magnitude of the gravitational force between an m on the surface of the Earth due to the Sun to that due to the Moon is 169.6.

Learn  more about gravitational force here: brainly.com/question/72250

#SPJ1

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Answer:

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A block of mass 9.5kg rests on a slope an angle of 23.0∘ relative to the horizontal. What is the size of the contact force norma
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The Normal Force = M x G x Cos(theta)

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3 years ago
a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th
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-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

  Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>

<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock.  A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

    His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock.  The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so 

                     (300 x speed) = 36 kg-m/s .

Divide each side by 300:  speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum.  The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed 
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
                                                         (300 x speed) = 186

Divide each side by 300:  speed = 186/300 = <em>0.62 m/s</em>    <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.

  yay !

By the way ... thanks for the 6 points.  The warm cloudy water
and crusty green bread are delicious.


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Lie scales are used when there are redundant questions that ask the same questions over and over with different formats to see if the answers are consistent over time.

Learn more about assessments at brainly.com/question/556621

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