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3 years ago

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The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 {\rm mA} for 37 continuous hours. During t

hat time the voltage will drop from 1.5 {\rm V} to 1.0 {\rm V} . Assume the drop in voltage is linear with time.Part A How much energy does the battery deliver in this 37h interval?Express your answer to two significant figures and include the appropriate units.

1 answer:

borishaifa [10]3 years ago

7 0

Answer:

1.42 KJ

Explanation:

solution:

power in beginning $p_{0}$ =(1.5 V).(9× $10^{-3}$ A)

= 13.5 mW

after continuous 37 hours it drops to

$p_{37}$ =(1 V).(9× $10^{-3}$ A)

=9 mW

When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.

37 hours= 37.60.60

=133200‬ s

w=(9× $10^{-3}$ A×133200‬ )+ $\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)$

=1.42 KJ

<em>NOTE:</em>

There maybe a calculation error but the method is correct.

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Answer:

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b).0.3011 m

c).0.0719 m

d).0.2137 N

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Explanation:

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Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

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Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

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b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

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Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

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d). Distance from the leading edge upto which the flow will be laminar,

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We know that the force acting on the plate is

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and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

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Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

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= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

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