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3 years ago

7

heat engine operating between 100 °C and 700 °C has an efficiency equal to 40% of the maximum theoretical efficiency. How much e

nergy does this engine extract from the hot reservoir in order to do 5000 J of mechanical work?

1 answer:

ipn [44]3 years ago

3 0

Answer:

Explanation:

Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.

= 600 / (273 + 700 )

= 600 / 973

= .6166

operating efficiency = 40% of .6166

= .4 x .6166

= .2466 = 24.66 %

efficiency = work output / heat input

= 5000 / heat input = .2466

heat input = 5000 / .2466

= 20275.75 J .

HEAT EXTRACED = 20275.75 J.

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3 years ago

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Answer:

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Explanation:

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Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

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Hence, the magnetic field is $B=2.958\times10^{-5}T$

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3 years ago

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scoray [572]

Answer:

<em>The initial charges on the spheres are </em> $6.796\ 10^{-6}\ c$ and $-3.898\ 10^{-6}\ c$

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

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$\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}$

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$\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]$

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

$\displaystyle q=\frac{q_1+q_2}{2}$

Using this second condition:

$\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756$

$\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C$

Solving for q2

$\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1$

Replacing in [1]

$\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}$

Rearranging, we have a second-degree equation for q1.

$\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0$

Solving, we have two possible solutions

$\displaystyle q_1=6,796.10^{-6}\ c$

$\displaystyle q_1=-3.898.10^{-6}\ c$

Which yields to two solutions for q2

$\displaystyle q_2=-3.898.10^{-6}\ c$

$\displaystyle q_2=6.796.10^{-6}\ c$

Regardless of their order, the initial charges on the spheres are $6.796\ 10^{-6}\ c$ and $-3.898\ 10^{-6}\ c$

8 0

3 years ago