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jolli1 [7]
3 years ago
6

powers the fan is converted to the kinetic energy of moving air. A fan is putting 1.9 J of kinetic energy into the air every sec

ond. Then the fan speed is increased by a factor of 2. Air moves through the fan faster, so the fan moves twice as much air at twice the speed.How much kinetic energy goes into the air every second?Express your answer with the appropriate units.
Physics
1 answer:
S_A_V [24]3 years ago
7 0

Answer:

Explanation:

Given

Power P=1.9\ J/s

Also Power is given by

P=\frac{\rho \times A\times V^3}{2}----1

Where P=Power delivered

\rho=density of air

A=Area

V=Velocity of air

If we increase velocity by a factor of 2 Power changes to P'

P'=\frac{\rho \times A\times (2V)^3}{2}----2

Divide 1 and 2

\frac{P}{P'}=\frac{V^3}{(2V)^3}

P'=8 P

P'=8\times 1.9=15.2\ J

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The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.
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Answer:

i = 2.84 \times 10^{-3} A

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

J = \frac{di}{dA}

so the current through the wire is given as

i = \int J dA

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here we have

J = (3.25 \times 10^8)r^2

now plug in the values in above equation

i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr

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i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr

i = (2.04 \times 10^9) \frac{r^4}{4}

now plug in both limits as mentioned

i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})

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3 years ago
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