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3 years ago

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In a physics laboratory experiment, a coil with 170 turns enclosing an area of 10.9 cm2 is rotated during the time interval 3.50

×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10−5 T. What is the total magnitude of the magnetic flux (initial) through the coil before it is rotated?

1 answer:

Leni [432]3 years ago

3 0

$N= 170 turns\\A=10.9cm^2\\t=3.5*10^{-2}s \\\B = 5.6*10^{-5}T$

The Magnetic flow $\Phi_{initial}$ is given by the formula,

$\Phi_{Initial}=BAsin\theta$

Replacing the values

$\Phi_{Initial} =(5.6*10^{-5})(10.9)(\frac{10^{-4}m^2}{1cm^2}) sin90\°$

$\Phi_{Initial} =6.104*10^{-7} Wb$

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4 years ago

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polet [3.4K]

Answer:8540 kg-g/s

Explanation:

Given

mass of blue car $m_c=427 kg$

velocity of blue car $v_c=20 m/s$

mass of the truck $m_t=1282 kg$

speed of truck $v_t=13 m/s$

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Let v be the velocity of combined system at angle \theta from vertical

Conserving momentum in east direction

$m_c\times v_c=(m_c+m_t)v\cos \theta$

$427\times 20 =1709\times v\cos \theta$ ------1

Conserving Momentum in Y direction

$m_t\times v_t=(m_c+m_t)v\sin \theta$

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squaring and then adding 1 & 2 we get

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4 years ago

Technician A states that a reading of 250 mv from the O2 indicates a rich exhaust. Technician B states that when the PCM receive

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Answer:

Technician B is correct

Explanation:

An oxygen sensor will generate about 1.0 volts when the fuel mixture is rich and there is little unburned oxygen in the exhaust. When the mixture is lean, the sensor's output voltage will drop down to about 0.1 volts.

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3 years ago

The circumference of a sphere was measured to be

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To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

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$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

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3 years ago

What is the efficiency of an engine that exhausts 440 J of heat to a cold reservoir and receives 570 J of heat from a hot reserv

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Answer:

Efficiency = 77%

Explanation:

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To find the efficiency;

$Efficiency = \frac {Out-put \; energy}{In-put \; energy} * 100$

Substituting into the equation, we have;

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Efficiency = 77.19 ≈ 77%

Therefore, the efficiency of the engine is 77 percent.

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3 years ago