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Marta_Voda [28]
3 years ago
11

Stephan Hawkins worked extensively with theoretical gravitational physics. He is well known for his work with black holes. He pa

ssed on March 14th, 2018. It has been observed that a certain star (S2) in the Sagittarius A* (meaning in the direction of the constellation Sagittarius) is orbiting a supermassive black hole. If the period of S2 is 16year and the distance between the two is 1.79e13m. What is the mass of this black hole? (be careful with your units)
Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
4 0

Answer:Watching the moving images and sounds in the video helped me picture what the text from A Black Hole Is NOT a Hole is describing about Centaurus A. Being able to both see and hear the information helped me gain a better understanding of black holes.

Explanation: bc am now

liubo4ka [24]3 years ago
3 0

Answer: M=1.3(10)^{34}kg

The equation that relates the period T of a body that orbits a greater body in space with the distance r  between both bodies is:

T^{2}=\frac{4\pi^{2}}{GM}r^{3} (1)  

Where;  

M is the mass of the Black Hole (the value we want to find)

G is the Gravitational Constant and its value is 6.674x10^{-11}\frac{m^{3}}{kgs^{2}}  

r=1.79(10)^{13}m is the distance from the Black Hole to the Star S2 (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit).  

T=16y is the orbital period of the Star S2

At this point, note we have to transform the units of T from years y to seconds:

T=16y.\frac{365days}{1y}.\frac{24h}{1day}.\frac{3600s}{1h}

T=504796000\approx 5(10)^{8}s is the orbital period of the Star S2 in seconds

If we want to find the mass M of the black hole, we have to express equation (1) as written below and substitute all the values:  

M=\frac{4\pi^{2}r^{3}}{G.T^{2}} (2)  

M=\frac{4\pi^{2}(1.79(10)^{13}m)^{3}}{(6.674x10^{-11}\frac{m^{3}}{kgs^{2}})(5×10^{8}s)^{2}} (3)  

M=\frac{4\pi^{2}5.73(10)^{39}m^{3}}{16675000\frac{m^{3}}{kg}} (3)  

Finally we have the mass of the black hole:

M=1.3(10)^{34}kg

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3 years ago
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Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
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How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?
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<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
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3 years ago
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We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a
o-na [289]

Answer:

The final temperature of the gas is <em>114.53°C</em>.

Explanation:

Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:

ΔU=Q - W

ΔU = 1180 J - 2020 J = -840 J

Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

ΔU=\frac{3}{2} nRΔT

ΔU=\frac{3}{2} nR(T_{2} -T_{1} )

Then we make the final temperature, T₂, subject of the formula:

T_{2} =\frac{2ΔU}{3nR} +T_{1}

T_{2} =\frac{2(-840J)}{(3)(5)(8.314J/mol.K)} +128 deg.C

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Therefore the final temperature of the gas, T₂, is 114.53°C.

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Answer: c is correct

Explanation: i did this

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