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Natasha_Volkova [10]
3 years ago
5

Explain the differences between 1- Energy 2- Power 3- Work 4- Heat Your answer should explain the mathematica and physical meani

ng. A thermodynamic example with a nice sketch for each case is a must.

Engineering
1 answer:
vazorg [7]3 years ago
4 0

Answer:

Explanation:

Energy : Energy can be defined as both qualitative and quantitative property that can be given to an object in the form of work or heat.

For example - water stored in a tank at a height,h from the ground has a potential energy of U = ρgh.

Mathematically, Energy of water stored in a tank is

                      U = ρ.g.h

where, ρ is  water density

            g is acceleration due to gravity

            h is height of water tank from ground

Power : Power is defined as rate at which work can be done.

For example - When a force F is applied to an object to move it through a distance d in time t seconds is an example of power delivered.

Mathematically, Power = Work done / Time

                         P = \frac{F\times  d}{t}

                                P = \frac{W}{t}

Work : When a force act on a body and the body moves, work is said to be done.

For example - Gas confined in a cylinder by a piston. The gas expands when it is heated, doing work on the piston.

Mathematically, Work,W = F X d

where, F is force

            d is distance

Heat : Heat is a form of energy which is produced as a result of motion of atoms and particles.

For example - while increasing the temperature of water, heat is added to the water by means of gas burner or any other form.

Mathematically we can calculate heat release or heat absorb by

    q = m.C.ΔT

where q is heat

           m is mass

           C is specific heat capacity

           ΔT is change in temperature

You might be interested in
Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en
nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows  

\eta =1-\dfrac{T_1}{T_2}

Here temperature should be in Kelvin.

For engine A

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{520+273}

\eta =0.63

For engine B

\eta =1-\dfrac{T_1}{T_2}

\eta =1-\dfrac{273+20}{820+273}

\eta =0.73

So from above we can say that engine B is more efficient.

4 0
3 years ago
Which of the following explains the main reason to cut a piece of wood on the outside of the measurement mark?
maks197457 [2]
I think it’s D ?? I’m not completely sure tho
4 0
3 years ago
At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing
Inessa [10]

Answer:

vB = - 0.176 m/s   (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒  (OA) = Sin 35°*(AB)

⇒   (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒  (OB) = Cos 35°*(AB)

⇒   (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒  (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒  vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒  vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒  vB = - 0.176 m/s   (↓-)

The pic can show the question.

7 0
3 years ago
Read 2 more answers
I want to solve the question
DedPeter [7]

Answer:

yes.

Explanation:

5 0
3 years ago
An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The
Mila [183]

Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

6 0
3 years ago
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