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2 years ago

Which of the choices describes an action–reaction force pair for a space station containing astronauts in orbit about the earth?

Physics

1 answer:

11111nata11111 [884]2 years ago

5 0

Answer:

(D) The weight of the space station and the gravitational force of the space station on the earth.

Explanation:

In both A and B , both the forces act in the same direction ( downwards ) , so they can not be action- reaction force .

In the option C , weight of a astronaut can only be reaction force of gravitational force exerted on the earth by astronaut. Both astronaut and the earth pull each other with equal and opposite force. So option D is correct.

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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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The given question is incomplete. The complete question is as follows.

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