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qwelly [4]
3 years ago
10

The Earth is instantly replaced in its orbit by a speck of dust. Which statement best describes the subsequent orbital motion of

that piece of dust? The Earth is instantly replaced in its orbit by a speck of dust. Which statement best describes the subsequent orbital motion of that piece of dust? The dust particle will continue in the same orbit as the Earth did, orbiting the Sun in 1 year. The dust particle will move to a smaller orbit and orbit the Sun in less than 1 year. The dust particle will move to a larger orbit and orbit the Sun in more than 1 year. The dust particle will spiral into the Sun. The dust particle will be ejected from the Solar System.
Physics
1 answer:
Alik [6]3 years ago
5 0

The correct statement is:

The dust particle will move to a larger orbit and orbit the Sun in more than 1 year.

In fact, the dust particle has smaller mass than the Earth, therefore the gravitational attraction exerted by the Sun on the dust will be smaller, according to the formula of the gravitational force:

F=G\frac{M_S m}{r^2}

where G is the gravitational constant, MS is the Sun mass, m is the mass of the Earth (at first) and then of the dust particle (later), and r is the distance from the Sun. When we replace the Earth with the dust particle we see that m decreases, so the gravitational force F decreases, and the particle will move to a larger orbit. Then its orbital period will increase, according to the third law of Kepler, which states that the square of the orbital period is proportional to the cube of the orbital radius:

T^2 \sim r^3

therefore, since the orbit is larger, the orbital period is greater.

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Natasha_Volkova [10]

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a stable star.

<h3>How does a protostar become a stable star?</h3>

The interstellar medium can sometimes be gathered into a large nebula, which is a cloud of gas and dust. A nebula can span a number of light years. These nebulae are where gas and dust can combine to produce stars. Until a star can combine hydrogen into helium, it cannot be considered a star. They are referred to as protostars before then. As gravity starts to gather the gases into a ball, a protostar is created. Accrution is the term for this procedure.

Gravitational energy starts to heat the gasses as gravity draws them into the ball's core, which causes the gasses to radiate radiation. Radiation initially just dissipates into space. However, much of the radiation is retained inside the protostar as it draws in stuff and becomes denser, which causes the protostar to heat up even more quickly.

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a star.

Learn more about a protostar here:

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3 0
1 year ago
HELP
Mila [183]

Answer: When you break on your bike and when you rub your hands together to get warm.

Explanation: Force and friction affect our daily lives in numerous amounts of ways. For instance, when a football is kicked, it moves faster later after some time its force decreases due to friction. A common example of friciton is when a bike stops. When the brakes are applied the friction on the pads cause the bike to stop. The rubbing hands is making friction. Which makes you get warm.

5 0
3 years ago
A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o
Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0
3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
3 years ago
Consider the cross section to the right with an overall height h 4.2 inch, bottom flange plate width b1-2.5 inch, top flange pla
ICE Princess25 [194]

Answer: a) 1.98inch

b) 2.737 2.44

c) 0.497inch

d) 1.58inch

Explanation:

Solution is attached in the pictures below

7 0
3 years ago
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