Ask question

Ask question

All categories

3 years ago

11

Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of

7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.

1 answer:

Triss [41]3 years ago

5 0

Answer:

4.408 $\mathsf{M_{sun}}$

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

$(m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}$

$\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}$

$\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}$

combined mass (m₁+m₂) = 4.408 $\mathsf{M_{sun}}$

You might be interested in

Define what is meant by restrictive interventions

zzz [600]

Medical movement for disabilities people

5 0

4 years ago

Which of the following is NOT a unit of measurement for speed?

Dmitriy789 [7]

I don't see any units of measurement or speed I don't see anything

3 0

3 years ago

Read 2 more answers

A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s

Nikolay [14]

Answer:

what if I do and b then someone else c I don't have enough time pls

8 0

3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago

This food chain is incomplete. Two groups of organisms are missing. One group of organisms that is missing are the producers, pl

Katena32 [7]

Answer:

Decomposers that recycle matter back into the food chain.

Explanation:

All dead organisms are food for decomposers. Decomposers that recycle matter back into the food chain.

5 0

4 years ago