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3 years ago

In each cycle, a heat engine performs 710 J of work and exhausts 1480 J of heat. What is the thermal efficiency?

Physics

1 answer:

Brrunno [24]3 years ago

5 0

Answer:

D) 0.32

Explanation:

The computation of the thermal efficiency is given below:

Thermal efficiency is

= 710 J ÷ (710 J + 1480 J)

= 710 J ÷ 2,190 J

= 0.32

Hence, the thermal efficiency is 0.32

Therefore the last option is correct

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Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the cur

olganol [36]

Answer:

To find the diameter of the wire, when the following are given:

Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.

Using the ohm's law,

Resistance R = (rho*L)/A

R = V/I.

Crossectional area of the wire A = π*square of radius

Radius = sqrt(A/π)

Diameter = Radius/2 = [sqrt(A/π)]

Making A the subject of the formular

A = (rho* L* I)V.

From the result of A, Diameter can be determined using

Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7

Explanation:

Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

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3 years ago

A college student is working on her physics homework in her dorm room. her room contains a total of 6.0×1026 gas molecules. as s

ella [17]

<span>6.6 degrees C Let's model the student as a 125 w furnace that's been operating for 11 minutes. So 125 w * 11 min = 125 kg*m^2/s^3 * 11 min * 60 s/min = 82500 kg*m^2/s^2 = 82500 Joule So the average kinetic energy increase of each gas molecule is 82500 J / 6.0x10^26 = 1.38x10^-22 J Now the equation that relates kinetic energy to temperature is: E = (3/2)Kb*Tk E = average kinetic energy of the gas particles Kb = Boltzmann constant (1.3806504Ă—10^-23 J/K) Tk = Kinetic temperature in Kelvins Notice the the energy level of the gas particles is linear with respect to temperature. So we don't care what the original temperature is, we just need to know by how much the average energy of the gas particles has increased by. So let's substitute the known values and solve for Tk E = (3/2)Kb*Tk 1.38x10^-22 J = (3/2)1.3806504Ă—10^-23 J/K * Tk 1.38x10^-22 J = 2.0709756x10^-23 J/K * Tk 6.64 K = Tk Rounding to 2 significant digits gives 6.6K. So the temperature in the room will increase by 6.6 degrees K or 6.6 degrees C, or 11.9 degrees F.</span>

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3 years ago

A mercury thermometer is constructed as

Tamiku [17]

Answer:

The change in height of the mercury is approximately 2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

$\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C$

from which we solved for the change in volume $\Delta V$ due to a given change in temperature $\Delta T$

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

$V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3$

Therefore, the change in volume with a change in temperature of 36°C becomes:

$\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3$

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

$V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm$

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3 years ago

The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 36t + 6 for t ≥ 0

igor_vitrenko [27]

1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:

$v(t)=s'(t)=(2t^3-3t^2-36t+6)'=6t^2-6t-36$

2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:

$a(t)=v'(t)=(6t^2-6t-36)'=12t-6$