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serg [7]
3 years ago
10

The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 x 10

6 m. What is the mass of planet Nutron
Physics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

4.3 * 10^28 kg

Explanation:

Given:

Period, T = 84s

Radius of satellite orbit, r = 8*10^6

Using the relation :

M = 4π²r³ / GT²

Where G = Gravitational constant, 6.67 * 10^-11

M = 4*π^2*(8*10^6)^3 / 6.67 * 10^-11 * 84^2

M = (20218.191872 * 10^18) / 47063.52 * 10^-11

M = 0.4295937 * 10^18 - (-11)

M = 0.4295937 * 10^29

M = 4.295937 * 10^28 kg

M = 4.3 * 10^28 kg

Mass of planet Nutron = 4.3 * 10^28 kg

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Explanation:

it can be safely concluded that an object moving in a circle at constant speed is indeed accelerating. It is accelerating because the direction of the velocity vector is changing.

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7 0
3 years ago
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 pis
Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0
3 years ago
A bird lands on a bare copper wire carrying a current of 51
nirvana33 [79]
The resistance of the piece of wire is
R= \frac{\rho L}{A}
where
\rho = 1.68 \cdot 10^{-8}\Omega m is the resistivity of the copper
L=5.1 cm=0.051 m is the length of the piece of wire
A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2 is the cross sectional area of the wire
By substituting these values, we find the value of R:
R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V
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How does the physical structure of a plant have<br> Large leaves help it survive?
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