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Contact [7]
3 years ago
12

A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is consider

ed to be stable and does not change over time. The student claims, "The only experimentally measurable external force exerted on the planet is the force due to gravity from the star." Is the student’s claim supported by the evidence? What reasoning either supports or contradicts the student’s claim?
Physics
2 answers:
bazaltina [42]3 years ago
5 0

Answer:

Explanation:

Matter warp the space.  

The first law of Kepler states that planets orbits sun, in an elliptical curve but in a plane, it is a very simple motion in fact. And according to Newton Gravitional force look like an snapshot force acting on bodies due to the presence of a very big mass body (like earth).

Special and General Relativity theory (Einstein) states speed of light as the maximum possible speed, and the way to look gravitional force is different. According to field theory, matter warp space, so objects have to keep moving along that deformation. It is noy possible snapshot forces, and there is only one way to go.

We rather have to say planets go through paths determine by gravity  

AleksAgata [21]3 years ago
5 0

Answer:

Yes, the student's claim is supported by the evidence.

There are different other forces acting on the planet but with insignificant values.

Explanation:

The student claimed that the gravitational force is the only measurable external force acting directly on the planet from the star and it has been proven experimentally. However, there are different other external forces acting on the planet and these forces have insignificant values and have not been supported by any experimental data.

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Zolol [24]

Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

T = 2\pi \sqrt{\frac{r^3}{GM} }

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

 

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N

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3 years ago
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As a type of thermal power station, a coal-fired power station converts chemical energy stored in coal successively into thermal energy, mechanical energy and, finally, electrical energy. The coal is usually pulverized and then burned in a pulverized coal-fired boiler.Coal-fired plants produce electricity by burning coal in a boiler to produce steam. The steam produced, under tremendous pressure, flows into a turbine, which spins a generator to create electricity.

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3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

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3 years ago
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