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Snezhnost [94]
2 years ago
10

In the equation for the gravitational force between two objects, which quantity must be squared?

Physics
1 answer:
bezimeni [28]2 years ago
8 0

Answer:

d

Explanation:

The quantity that must be squared in the equation of gravitational force is distance d.

According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.

  • Therefore, the quantity to be squared is d

The formula is given as:

  Fg  = \frac{G m_{1} m_{2} }{d^{2} }  

So d is the quantity that must be squared

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A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastwar
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<h2>a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    \texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m

    Magnitude of his displacement = 2729.47 m

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The answer would be melting point.

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In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed
Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is 87.87 * 10^{-37}.

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= 5.29 * 10^{-11} m\\

velocity=2.9* 10^{6} m/s

Working formula, M=N/A

I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}

T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }

   =15.16 * 10^{-5} s

I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }= 0.10 * 10^{-14}

                     =1 * 10^{-15} C

M=1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}

  =87.87 * 10^{-37}

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

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