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Reptile [31]
2 years ago
12

A Carnot engine has an efficiency of 83.0% and performs 4500 J of work every cycle. How much energy is discharged to the lower t

emperature reservoir every cycle?
Physics
1 answer:
ololo11 [35]2 years ago
6 0

We will start by finding the heat in state one through the thermal efficiency or efficiency of a thermal machine is a coefficient or dimensionless ratio calculated as the ratio of the energy produced (in an operating cycle) and the energy supplied to the machine. Mathematically it is the relationship between the work generated and the heat emanated.

So,

\eta = \frac{W}{Q_1} \rightarrow Q_1 = \frac{W}{\eta}

Q_1 = \frac{4500}{0.83}

Q_1 = 5421J

The total change of energy is equivalent to 4500J and this is equal by conservation of energy to the total change in heat. So:

W = \Delta Q

W = Q_1-Q_2

4500 = 5421- Q_2

Q_2 = 921.68 J

Therefore the energy which is discharged to the lower temperature reservoir every cicle is 921.7J

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Abnormal protrusion of the eye out of the orbit is known as
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Answer:

Exophthalmos

Explanation:

Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.

It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.

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7 0
3 years ago
A heat engine with a thermal efficiency of 45 percent rejects 500 kj/kg of heat. how much heat does it receive
adell [148]
Alot as far as i know unless you need it in formal terms.
8 0
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select the example that best describes a renewable resource. a. all the houses built in the new neighborhood had natural gas ava
Bad White [126]
B. my town is powered by electricity that is generated by the energy from the flow of water through a large dam.

The water from the large dam is example of renewable energy. It can be replenished through rainfall cycle, so it is a renewable form of energy. 
4 0
3 years ago
Read 2 more answers
an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr
pantera1 [17]

The horizontal velocity was constant, so:

s = vt

s = 10\cdot9

s = 90

it traveled 90meters

6 0
3 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
3 years ago
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