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2 years ago

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An electric motor consumes 10.8 kJ of electrical energy in 1.00 min . Part A If one-third of this energy goes into heat and othe

r forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 rpm

1 answer:

lina2011 [118]2 years ago

8 0

Answer:

0.458 N/m

Explanation:

Power = Energy/time

From the question,

Note: two-third of the energy into the motor,

Therefore,

P = 2/3(E)/t................. Equation 1

Where P = power of the motor, E = Electrical Energy, t = time

Given: E = 10.8 kJ = 10800 J, t = 1.00 min = 1/60 s

Substitute into equation 1

P = 10800(1/60)(2/3)

P = 120 W.

But,

The power of a rotating motor is given as,

P = Tω................ Equation 2

Where T = Torque of the engine, ω = angular velocity of the engine

Make T the subject of the equation

T = P/ω............... Equation 3

Given: P = 120 W, ω = 2500 rpm = (2500×0.1047) rad/s = 261.75 rad/s

Substitute into equation 3

T = 120/261.75

T = 0.458 N/m

Then the torque develop in the engine = 0.458 N/m

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<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>

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2 years ago

Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme

Artyom0805 [142]

Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider $T_{1}\ and\ T_{2}$ are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

$Let\ Q_{1}\ and\ Q_{2}$ are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

$W\ =\ Q_{1}-Q_{2}$

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

$Efficiency\ [\eta]=\frac{W}{Q_{1}}$

$\eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}$

$=\ 1-\frac{Q_{2}} {Q_{1}}$

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

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2 years ago

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A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

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2 years ago

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val

OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container is 0.5056

The Total Heat transfer is $Q_{in} = 22.62 KJ$

Explanation:

Explanation is in the following attachments

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Art [367]

Answer:

The total work done is 5997.6 J

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As per the question:

Mass of the bag, m = 60 kg

Vertical distance, h = 9 m

Mass lost, m' = 12 kg

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Lost mass is proportional to the square root of the distance covered while lifting:

m' ∝ $\sqrt{h}$

m' = $K\sqrt{9}$

where

K = proportionality constant

12 = 3K

K = 4

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$m(h) = 60 - k\sqrt{h}$

Work done is given by:

$W = \int_{0}^{h}m(h)gdh$

$W = \int_{0}^{9}(60 - k\sqrt{h})gdh$

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$W = 9.8\times (540 + \frac{8}{3}\times 27) = 5997.6\ J = 5.9976\ kJ$

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2 years ago

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