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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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510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr
Yuki888 [10]

Answer:

The terminal velocity is v_t  =17.5 \ m/s

Explanation:

From the question we are told that

       The mass of the squirrel is  m_s  =  50\ g  =  \frac{50}{1000} =  0.05 \  kg

      The surface area is   A_s =  935 cm^2  =  \frac{935}{10000} = 0.0935 \ m^2

       The height of fall is  h =4.8 m

        The length of the prism is l =  23.2 = 0.232 \ m

          The width of the prism is w =  11.6 =  0.116 \ m

 

The terminal velocity is mathematically represented as

       v_t  =  \sqrt{\frac{2 * m_s *  g }{\dho_s * C  * A } }

Where \rho  is the density of a rectangular prism with a constant values of \rho  =  1.21 \ kg/m^3

            C is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      A =  0.116 * 0.232

       A =  0.026912 \ m^2

 substituting values

      v_t  =  \sqrt{\frac{2 * 0.510 *  9.8 }{1.21 * 1  * 0.026912 } }

     v_t  =17.5 \ m/s

       

7 0
4 years ago
A monatomic ideal gas has pressure p1 and temperature T1. It is contained in a cylinder of volume V1 with a movable piston, so t
Vanyuwa [196]

Answer:

A) Q1 = (3/2)P1V1[A - 1]

B) W2 = P1V1(In A)

C) W3 = P1V1(1 - A)

Explanation:

A) From first law of thermodynamics and applying to the question, we have;

ΔU = Q - W

Where,

ΔU = change in internal energy

Q = the heat absorbed

W = the work done

Now, because the first process occurs at constant volume, the work done is zero:

Thus,

ΔU = Q - 0

ΔU = Q

The change in internal energy is given by;

ΔU = nCvΔt

where;

n = the number of moles of the gas

R = the gas constant,

Cv = the specific heat at constant volume

Δt = The change in temperature i.e T2 - T1.

Now, using the ideal gas law, let us find an expression for n and Δt

P1V1 = nRT1

n = P1V1/RT1

T1 = P1V1/nR

Now, the specific heat at constant volume is Cv = (3/2)R

Now, from the question, since it's pressure has reached AP1, we can calculate the temperature T2 by using the ideal gas law at the new conditions of the gas as;

AP1V1 = nRT2

T2 = AP1 V1/ nR

Now, we are to express the heat added in terms of p1, V1, and A

Q = ΔU = nCv(T2 - T1)

From earlier, we saw that,

T1 = P1V1/nR

Putting equation of T2 and T1 into the energy equation to get;

Q = nCv((AP1 V1/ nR) - P1V1/nR)

Q = Cv • P1V1/R (A - 1)

Now, from earlier, we saw that Cv = (3/2)R. Thus,

Q = (3/2)R • P1V1/R (A - 1)

Q = (3/2)P1V1[A - 1]

B) Here again, we are to express work done in step 2 in terms of p1, V1, and A.

This process is an isothermal process because temperature is constant and so work done is given as; W = nRT In(V2/V1)

T = T1 because temperature is constant

From earlier, we saw that;

n = P1V1/RT1 and

But in this process, it's

n = P1V1/RT1 and thus,

V2 = nRT2/P1

We also saw that T2 = AP1 V1/ nR

V1 = nRT2/AP1

Plugging in the relevant values into, W = nRT In(V2/V1), we obtain;

W = (P1V1/RT1) • RT1 • In((nRT2/P1)/(nRT2/AP1)

W = P1V1(In A)

C) In step 3,we have and isobaric process because the pressure is constant.

Work done in this case is given by ;

W = P(V1 - V2)

Because V2 in now the final volume while V1 is now the the initial volume

Now, P is P1 because it's an isobaric process.

From earlier, we saw that,

V1 = nRT2/AP1 and V2 = nRT2/P1

And that T2 = AP1 V1/ nR

Thus,

V1 = V1 and V2 = AV1

Thus, W = P1(V1 - AV1) = P1V1(1 - A)

4 0
3 years ago
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n200080 [17]

Answer:

.....

Explanation:

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makkiz [27]
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