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madreJ [45]
3 years ago
9

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

ther step to isolate it. Let's say we now have
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers.
Physics
1 answer:
Margaret [11]3 years ago
5 0

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

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a) the mass of the star Alpha is M_{\alpha}= 7.80*10^{29} kg

b) the mass of the star Beta is M_{\beta} = 2.34*10^{30}kg

c)  the radius of the orbit of the orange star R_{0}= 1.9*10^{9}m

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e) the orbital speed of the orange star V_{0} = 4.4*10^{2}km/s

f)  the orbital speed of the black hole V_{B} =77 km/s

Explanation:

The generally formula for orbital speed is given as v =\frac{2\pi R}{T}

    where v is the orbital speed

                R  is the radius of the star

               T  is  the orbital period

From the question we are given that

  alpha star has an orbital speed of  V_{\alpha} = 36km/s = 36000m/s

  beta star has an orbital speed of V_{\beta} = 12km/s = 12000m/s

   the orbital period is  T = 137d = 137(86400)sec   1 day is equal 86400 seconds

  Making R the subject of formula  we have   for the radius of the alpha star as

 

                 R_{\alpha} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(36000m/s)}{2\pi}

                       =6.78*10^{10}m

   for the radius of the Beta star as

                 R_{\beta} =\frac{V_{\alpha}T}{2\pi} =\frac{2*(86400s)*137*(12000m/s)}{2\pi}

                                 = 2.26*10^{10}m

Looking at the value obtained for R_{\alpha} and R_{\beta}

Generally the moment about the center of the mass are equal then

         M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta}

Thus 3M_{\alpha} = M_{\beta} -------(1)

Generally the formula for the orbital period is given as

                  T =\frac{2\pi(R_{\alpha+R_{\beta}})^{3/2 }}{\sqrt{G(M_{\alpha}+M_{\beta})} }

Then

        M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}

Where G is the gravitational constant given as 6.67408 × 10^{-11} m^3 kg^{-1} s^{-2}

        M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(6.78*10^{10}m +2.26*10^{10}m)^{2}}{(137d*86400s/d)^{2}(6.67*10^{-11}m^{3}kg^{-1}s^{-2})}

       M_{\alpha} + M_{\beta} = 3.12*10^{30}kg -------(2)

Thus solving (1) and (2) equations

    Mass of alpha star is  M_{\alpha}=7.80*10^{29}kg

and the Mass of Beta is M_{\beta} =2.34*10^{30}kg

Considering the equation

                     M_{\alpha}+M_{\beta} =\frac{4\pi^{2}(R_{\alpha}+R_{\beta})^{3}}{T^{2}G}

Making R_{\alpha} + R_{\beta} the subject

                        R_{\alpha}+ R_{\beta} =\sqrt[3]{\frac{(M_{\alpha+M_{\beta}})T^{2}G}{4\pi^{2}} } -------(3)

and considering this equation  M_{\alpha} R_{\alpha} = M_{\beta}R_{\beta} from above

               we have that R_{\beta} =\frac{M_{\alpha}R_{\alpha}}{M_{\beta}}

  Considering Question C

Let the Orange star be denoted by (0) and

Let the black-hole be denoted by (B)

And we are told from the question that

    Mass of orange star M_{0} = 0.67M_{sun} and

    Mass of black hole  M_{B} =3.8M_{sun}

And mass of sun is M_{sun}  = 1.99*10^{30}kg

Then R_{B} = [\frac{0.67M_{sun}}{3.8M_{sun}} ]R_{0} =0.176R_{0}--------(4)

We are also given that the period is T =7.75 days = 7.75 (86400s)

Considering equation 3

   

R_{0} + 0.176R_{0} = \sqrt[3]{\frac{(0.67+3.8)(1.99*10^{30}kg)(7.75(86400s))^{2} 6.67*10^{-11}Nm^{2}/s^{2}}{4\pi ^{2}} }

     Thus for V616 Monocerotis, R_{0} =1.9*10^{9}m

Considering  equation 4

The black-hole is

     R_{B}= 0.176R_{0}= 0.176*1.9*10^9 =34*10^8m

From the formula for velocity of  V_{0} = \frac{2\pi R_{0}}{T}  = 4.4*10^{9}km/s

         the velocity of V_{B} = \frac{2\pi R_{B}}{T} =77km/s

8 0
3 years ago
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