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miskamm [114]
3 years ago
6

2/25/20 or 2/28/20 Dispatch #53

Physics
1 answer:
mixer [17]3 years ago
3 0

Answer:

Power = 21[W]

Explanation:

Initial data:

F = 35[N]

d = 18[m]

In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.

Therefore:

Work = W = F*d = 35*18 = 630 [J]

And power is defined as the amount of work performed in a time interval.

Power = Work / time

Time = t = 30[s]

Power = 630/30

Power = 21 [W]

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Please help i forgot how to solve this, thank you.
mixas84 [53]

Answer:

The radius of its orbit is R \approx 1.899\times 10^{9}\,m.

Explanation:

Let suppose that Callisto rotates around Jupiter in a circular path and at constant speed, then we understand that net acceleration of this satellite is equal to the centripetal acceleration due to gravity of Jupiter. That is:

\omega^{2}\cdot R = a_{net} (1)

Where:

\omega - Angular speed, measured in radians per second.

R - Radius of the orbit, measured in meters.

a_{net} - Net acceleration, measured in meters per square second.

In addition, angular speed can be described in terms of period (T), measured in seconds:

\omega = \frac{2\pi}{T} (2)

And the net acceleration by the Newton's Law of Gravitation:

a_{net} = \frac{G\cdot m}{R^{2}} (3)

Where:

G - Gravitation constant, measured in cubic meters per kilogram-square second.

m - Mass of Jupiter, measured in kilograms.

Now we apply (2) and (3) in (1) to derive an expression for the radius of the orbit:

\frac{4\pi^{2}\cdot R}{T^{2}} = \frac{G\cdot m}{R^{2}}

R^{3} = \frac{G\cdot m \cdot T^{2}}{4\pi^{2}}

R = \sqrt[3]{\frac{G\cdot m\cdot T}{4\pi^{2}} } (4)

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, m = 1.90\times 10^{27}\,kg and T = 1460160\,s, then the radius of the orbit of Callisto is:

R = \sqrt[3]{\frac{(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} )\cdot (1.90\times 10^{27}\,kg)\cdot (1460160\,s)^{2}}{4\pi^{2}} }

R \approx 1.899\times 10^{9}\,m

8 0
2 years ago
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