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kvv77 [185]
3 years ago
13

A particular engine has a power output of 5 kW and an efficiency of 30%. If the engine expels 6464 J of thermal energy in each c

ycle, find the heat absorbed in each cycle. Answer in units of J.
Physics
1 answer:
Lorico [155]3 years ago
4 0

Answer:

The heat absorbed in each cycle is 9,234.286 J

Explanation:

Given;

power output, P = 5 kW = 5,000 W

efficiency of the engine, e = 30 % = 0.3

thermal heat expelled, Q_c = 6464 J

let the heat absorbed = Q_h

The efficiency of the engine is given as;

e = \frac{W}{Q_h} = \frac{Q_h-Q_c}{Q_h} = \frac{Q_h}{Q_h} - \frac{Q_c}{Q_h}  = 1-\frac{Q_c}{Q_h}\\\\e = 1-\frac{Q_c}{Q_h}\\\\0.3 = 1-\frac{Q_c}{Q_h}\\\\\frac{Q_c}{Q_h} = 1-0.3\\\\\frac{Q_c}{Q_h} = 0.7\\\\Q_h = \frac{Q_c}{0.7} \\\\Q_h = \frac{6464}{0.7} = 9,234.286 \ J.

Therefore, the heat absorbed in each cycle is 9,234.286 J.

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Answer:

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Explanation:

#Consider a circular area of radius R=2.98cm in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of B=0.23T and points toward the +ve z-axis. The angle between the magnetic field and the area is \theta=0. Hence the magnetic flux:-

\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2

Hence flux magnitude in +z direction is 3.0528\times10^-^3T \ m^2

b. We now find the magnetic flux when the field has a magnitude of <em>B=0.230T</em> and points at an angle of \theta=53.1\textdegree from the +z direction.

Magnetic flux is calculated as:

\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2

Hence the flux at an angle of 53.1\textdegree is 1.83299\times 10^-^3T \ m^2

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0

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The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

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The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

\theta = (\frac{\omega_0 + \omega}{2})t

where:

\theta is the angular displacement of the cd during the time interval t

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Here we have:

t = 1.36 s

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Therefore, the angular displacement of the cd during this time is:

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And since 1 rev = 2 \pi rad, we can convert into number of revolutions completed:

\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev

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