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2 years ago

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Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

ic force in the earth’s electric field, which is typically 100 N/C, directed downward.
a) What is the ratio of the electric force on the bee to the bee's weight?
b) What electric field strength would allow the bee to hang suspended in the air? Express your answer with the appropriate units.

1 answer:

Sedbober [7]2 years ago

4 0

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

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2.45 Kg

Explanation:

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Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

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Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

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So, let's isolate $T_{out}$ :

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Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

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$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

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$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
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