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Solnce55 [7]
3 years ago
9

What would be the weight of the moon if it were resting on the surface of the earth

Physics
1 answer:
kari74 [83]3 years ago
7 0
We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers. 
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).

Earth's radius . . . . .  6,360 km  =  6.36 x 10⁶ meters
Moon's radius . . . . .  1,738 km  =  1.738 x 10⁶ meters
Sum of their radii  =                      8.098 x 10⁶ meters

Also:
Earth's mass . . . . .  5.972 x 10²⁴ kg
Moon's mass . . . . .  7.348 x 10²²  kg
<span>
and now we're ready to go !

       Gravitational force = 

                   G  M₁ M₂ / R²

= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²²  kg)/</span>(8.098 x 10⁶ m)²

= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³)      Newtons

=    (I get ...)        4.463 x 10²³ Newtons

That's almost exactly   10²³ pounds 

                           =  50,153,000,000,000,000,000 tons.     

Those are big numbers. 
All I can say is:  I wouldn't exactly call that "resting" on the surface".
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Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
3 years ago
If a bullet travels at 593.0 m/s, what is its speed in miles per hour?
Ksenya-84 [330]

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1 \; hour = 3600 \; s. So 1 \; s = \frac{1}{3600}  \; hour.

Thus we can convert the units of the given quantity.

That is,

593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\&#10;593\;m/s=1,326.51\;miles/hour.

The quantity is converted to the required units.


7 0
3 years ago
A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an
BARSIC [14]

Answer:

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Explanation:

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Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

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Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

F=140*9.81=1,373.4 N

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6 0
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Many various units for volume are used but most often used unit is liter. In a metric system basic units are those such as meter, kilogram and liter while in imperial system units used are those such as foote, inch, pound and gallon.

Unit for volume in metric system is cubic meter. It is equal to a volume of a cube whose all sides measure 1m. This is equal to 1000L. For watering cans that contain several liters units used is decimeter cubed. 1dm^3 = 1L
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mash [69]

Answer:

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Hope I get a brainliest answer.

8 0
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