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vredina [299]
3 years ago
13

Answer the following three questions in complete sentences. 1. How much does the earth weigh? 2. How far away is the sun? 3. Wha

t is a black hole?
Physics
2 answers:
Black_prince [1.1K]3 years ago
7 0

Answer:

The Earth weighs 5.972 × 10^24 kg

The Sun is 94.022 million miles away.

A black hole is is a region of spacetime where gravity is so strong that nothing, no particles or even electromagnetic radiation such as light can escape from it. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.

I hope this helps!! :)

Effectus [21]3 years ago
6 0

Answer: find the answer in the explanation.

Explanation:

1.) The weight of the earth can be calculated and estimated by the physicist by considering the mass of the earth and the knowlegde of gravity.

If the mass of the earth is considered to be 6 × 10^24 kg. Then with Physics formula and some mathematical techniques, we can conclude that the weight of the earth is approximately equal to 6 × 10^25 N.

2.) The distance of the sun to the earth is approximately 1.5 × 10^8 km. This measure is the distance from the surface of the earth to the sun.

3.) The black hole is approximately the size of the earth of very large gravitational field strength. It is a region of space of enormous gravity where no particular object can escape from it. Even electromagnetic radiation or light cannot escape from it.

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Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a
Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }

Sum of the potential at point p is

V = Va + Vb

that is

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]

the expression below can be written as the equivalent

\frac{1}{(a^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }

likewise,

\frac{1}{(b^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }

hence,

V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

by reciprocal rule

1/a² = a⁻²

V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]

by binomial expansion of fractional powers

where (1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...

if we expand the expression we have the equivalent as shown

{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )

also,

{(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )

the above equation becomes

V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]

V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )

V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )

Answer

V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }

OR

V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }

8 0
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Harrizon [31]

Answer:

cartilage

Explanation:

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Give at least one fact about subatomic particles
victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

- These particles can be divided as Brayons and Leptons

- These particles are often held together by one of the four fundamental particles ( Weak force, strong force, electromagnetic force, gravitational force).

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2 years ago
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