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ruslelena [56]
3 years ago
7

A mass of 15 kg of air in a piston-cylinder device is heated from 25 o C to 77 o C by passing current through a resistance heate

r inside the cylinder. The pressure inside the cylinder is help constant at 300 kPa during this process and a heat loss of 60 kJ occurs. Calculate the electric energy supplied in kWh.
Physics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

The electrical energy supplied is 0.233 kWh

Explanation:

Given :

Mass of air m = 15 kg

Initial temperature T _{1} = 298 K

Final temperature T_{2} = 350 K

Heat loss Q _{out} = 60 kJ

From the first law of thermodynamics,

     \Delta U = W_{in} - W_{out} - Q_{out}

We know that internal energy is proportional to difference of temperature,

    W_{in} = m (T_{2} - T_{1}  ) + 60

    W_{in} = 15 \times  (350-298) + 60 }

Now we need answer into kWh so ( 1 kWh = 3600 kJ ).

So we need to multiply ( \frac{1}{3600} )

     W_{in} = [15 \times  (350-298) + 60 ] \times \frac{1}{3600} kWh

     W_{in} = 0.233 kWh

Therefore, the electrical energy supplied 0.233 kWh

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Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

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\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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+ 1.58 e -15

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