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kykrilka [37]
3 years ago
6

In a position vs. time graph depicting the motion of two different objects, the point at which the lines intersect is where the

objects:
a.
have the same position.
c.
collide.
b.
have the same speed.
d.
none of the above.
Physics
1 answer:
myrzilka [38]3 years ago
3 0
I think the correct answer from the choices listed above is option A. In a position vs. time graph depicting the motion of two different objects, the point at which the lines intersect is where the objects <span>have the same position. Hope this answers the question. Have a nice day.</span>
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7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin
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The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

\rm P_I(Initial pressure)=60 mpa = 60 bar

\rm  T_i(Initial temperature)=600° C

\rm P_e (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e

The change in the entropy is;

\rm S_2-S_1=\frac{\delta q}{dt}

The work done by the turbine is;

\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\  W_t =708.2 \ kJ/kg

Hence,the work done by the turbine will be 708.2 kJ/kg.

To learn more about the temperature, refer to the link;

brainly.com/question/7510619

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Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g
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Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

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v₂ = (0.006 m³/s)/(0.0000196 m²)

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