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ELEN [110]
2 years ago
11

A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 41,200 ly (light-years) from the galactic ce

nter and has a speed of 275 km/s. Estimate the total mass of the galaxy based on the motion of the star.
Gravitational constant is 6.674×10−11 m3/(kg·s2) and mass of the Sun Ms=1.99 × 1030 kg.
*Answer in billion solar mass
Physics
1 answer:
Ronch [10]2 years ago
6 0

The total mass of the galaxy is 443.4 Solar mass

Orbital velocity (v) = \sqrt{\frac{MG}{R} }

where M= weight of galaxy

G= gravitational constatnt = 6.674*10^-^1^1 (given)

R = distance from centre = 41200 Light years (given)= 4.12*9.5*10^1^6  km (1 ly= 9.5*10^3 billion km)

v= orbital velocity = 275  km/s (given)

∴ According to the formula

(2.75*10^2)^2 = \frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}

⇒ 7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1 (cross multiplying and expanding)

⇒ 29.59*10^2^1=M*6.674*10^-^1^1

⇒ \frac{29.59*10^2^1*10^1^1}{6.674}=M

⇒ 4.434*10^3^2=M

1 solar mass = 1.989*10^3^0 kg

⇒ Mass in solar mass =443.4 Solar mass

⇒ M = 443.4 Solar mass

Learn more about Orbital velocity here :

brainly.com/question/22247460

#SPJ10

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DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti
ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = 6.31\times 10^{4} T

Solution:

To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}

N = 22.74 = 23 turns (approx)

 

8 0
2 years ago
An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a
Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = 3.1 m

Width of the window = 2.1 m

Area of the window = (3.1\times 2.1) = 6.51\ m^2

Difference in air pressure = Inside pressure - Outside pressure

                                           = (1.0-0.954) atm = 0.046 atm

Conversion of the pressure in its SI unit.

⇒  1 atm = 101325 Pa

⇒ 0.046 atm = 0.046\times 101325 =4660.95 Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ Pressure=\frac{Force }{Area}

⇒ Force =Pressure\times Area

⇒ Plugging the values.

⇒ Force =4660.95\times 6.51

⇒ Force=30342.78 Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0
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MariettaO [177]

Explanation:

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Which would likely demagnetize a magnetized iron nail?
olga2289 [7]
The answer to this question is dropping it on a hard surface.
8 0
3 years ago
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