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Gnom [1K]
3 years ago
12

A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I

f the magnitude of the gravitational acceleration on the surface of the planet is 20.0 m/s2, what is the radius of the planet?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = \frac{4\times \pi^2\times r^3}{GM} where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

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Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

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This weight will act at the midpoint of the ladder

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An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

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Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0
3 years ago
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3. What is the acceleration of a 10 kg mass pushed by a 5 N force?
insens350 [35]

Answer:

F=ma

Plug it in:

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If a 7 kg bowling ball is lifted 2 m into the air and dropped, what speed will it strike the ground? (
Naily [24]

Answer:

6.32m/s

Explanation:

note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²

okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!

when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)

now enough of the talking let solve....

so the ball was dropped .ie from rest to the ground through a distance of 2m,

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solving ....

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V=u+at ,but since u=0

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therefore the speed the body uses to strike the ground is 6.32m/s

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