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3 years ago

Which of the following accurately describe some aspect of gravitational waves? Select all that apply.

Physics

1 answer:

steposvetlana [31]3 years ago

5 0

<h2>Answers:</h2>

-The first direct detection of gravitational waves came in 2015

-The existence of gravitational waves is predicted by Einstein's general theory of relativity

-Gravitational waves carry energy away from their sources of emission

<h2>Explanation:</h2>

Gravitational waves were discovered (theoretically) by Albert Einstein in 1916 and "observed" for the first time in direct form in 2015 (although the results were published in 2016).

These gravitational waves are fluctuations or disturbances of space-time produced by a massive accelerated body, modifying the distances and the dimensions of objects in an imperceptible way.

In this context, an excellent example is the system of two neutron stars that orbit high speeds, producing a deformation that propagates like a wave,<u> in the same way as when a stone is thrown into the water</u>. So, in this sense, gravitational waves carry energy away from their sources .

Therefore, the correct options are D, E and F.

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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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4 years ago

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3 questions I WILL MARK BRAINLYEST 100 POINTS!!!!!!

sammy [17]

Answer:

19 on the first one

Explanation:

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3 years ago

What do scientist do to receive suggestions an criticism of their research from other scientists?

barxatty [35]

They PUBLISH !

They provide complete written, detailed, technical descriptions of their thoughts, hypotheses, experiments, data, and conclusions, in publications that are read by other scientists around the world.

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3 years ago

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Boat A and Boat B have the same mass. Boat A’s velocity is three times greater than that of Boat B. Compared to the kinetic ener

Helen [10]

Answer:

Explanation:

D. Nine times as much.

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3 years ago

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A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is co

Lilit [14]

Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

8 0

3 years ago