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3 years ago

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What is the radius of the event horizon for a black hole with a mass 7.5 times the mass of the sun? This distance is called the

Schwarzschild radius. Express your answer with the appropriate units.

1 answer:

Illusion [34]3 years ago

4 0

Answer:

$1.1\times10^{4}m$

Explanation:

The Schwarzschild radius can be calculated as follows:

$R=\frac{GM}{c^2}$

Where, <em>G</em> is the gravitational constant, <em>M</em> is the mass and <em>c</em> is the speed of light.

Mass of sun is 1.98×10³⁰ kg

$R=\frac{G\times 7.5M_{sun}}{c^2}\\ R=\frac{6.67\times10^{-11}\times(7.5\times1.98\times10^{30})}{(3\times10^8)^2}\\ \Rightarrow R=1.1\times10^{4}m$

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Answer:

its a, c, d, and f

Explanation:

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When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d

Alchen [17]

Answer:

g = 0.85 m $s^{-2}$

Explanation:

g = $\frac{GM}{h^{2} }$

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x $10^{-11}$ N $m^{2}$ $kg^{-2}$ ), M is the mass of the earth (5.972 x $10^{24}$ kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

= 3.40 x 6371 km

h = 21661.4 km

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Thus,

g = $\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }$

= $\frac{3.9857 *10^{14} }{4.6922*10^{14} }$

= 0.84944

g = 0.85 m $s^{-2}$

The acceleration due to the Earth's gravitation is 0.85 m $s^{-2}$ .

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If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will Group of answer c

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Answer:

The correct answer is Option A (decrease).

Explanation:

According to Heisenberg's presumption of unpredictability, it's impossible to ascertain a quantum state viewpoint as well as momentum throughout tandem.
Also, unless we have accurate estimations throughout the situation, we will have a decreased consistency throughout the velocity as well as vice versa though too.

Other given choices are not connected to the given query. Thus the above is the right answer.

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A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

Therefore, the value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

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Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration

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Answer:

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