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zysi [14]
3 years ago
8

Forces always act in pairs? True or false

Physics
1 answer:
nasty-shy [4]3 years ago
7 0

Answer:

True

Explanation:

Just as Isaac Newton says, "For every action, there is an equal and opposite reaction."

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List at least three things that can happen as plate spread apart.
Nookie1986 [14]
Centre of Mass then axis of rotation and then moment of inertia. This was the toughest question for your level... happy to help ^_^. It was purely experimental question.
8 0
3 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
True or False. Centripetal force holds an object towards the center of a circle.
Flura [38]
False’ because it is a force that makes a body follow a curved path
7 0
2 years ago
TRUE OR FALSE! PLZ HELP
Ksju [112]

Answer:

True

Explanation:

Magnitude is the "value" the greater the value the greater the force is and vice versa

5 0
2 years ago
Suppose two objects are gravitationally attracted to each other with some force F. If the mass of object 1 is multiplied by a fa
RoseWind [281]

If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is

F' = G (5m₁) (2m₂) / d²

F' = 10  Gm₁m₂ / d²

but this is really just F' = 10F. So J is the correct choice.

8 0
2 years ago
Read 2 more answers
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