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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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In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of m
lara31 [8.8K]

Answer:

F=3.61\times 10^{-47}\ N

Explanation:

Mass of a proton, m_p=1.67\times 10^{-27}\ kg

Mass of an electron, m_e=9.11\times 10^{-31}\ kg

The distance between the electron and the proton is, r=5.3\times 10^{-11}\ m

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

F=G\dfrac{m_em_p}{r^2}

Where G is the universal Gravitational constant

F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N

So, the force between the electron and proton is 3.61\times 10^{-47}\ N.

7 0
3 years ago
A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from
Naddik [55]

Complete Question

The complete question iws shown on the first uploaded image  

Answer:

a

    y \to z \to x

b

  x \to z \to y

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

 Now the magnetic force is mathematically represented as

             F =  I L x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

 The motion of the loop is  from   y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z  

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

   The motion of the induced current will now be   x to z to y

 

5 0
3 years ago
What are two processes that transfer water into the atmosphere
Ira Lisetskai [31]
Water enters the atmosphere through evaporation, transpiration, excretion and sublimation: Transpiration is the loss of water from plant

Hope this helped you! :D
4 0
3 years ago
Q= Which one of the following statement is incorrect?
Lunna [17]

Answer:

A.

Explanation:

I think it might be the big number A

8 0
3 years ago
A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle i
olga55 [171]

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

6 0
3 years ago
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