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3 years ago

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A girl flies a kite at a height of 500 ft, the wind carrying the kite horizontally away from her at a rate of 20 ft/sec. How fas

t must she let out the string when the kite is 700 ft away from her? ft/sec

1 answer:

olya-2409 [2.1K]3 years ago

3 0

Answer:

she must let out the string when the kite is 700 ft away from her at 20 ft/s

Explanation:

given information:

a = the distance between the girl

b = the length of the string

a = 300

da/dt = 20 ft/s

c =500 ft, dc/dt = ?

Pythagorean Theorem,

$a^{2} + 300^{2}= c^{2}$

when y = 500

$a^{2} + 300^{2}= 500^{2}$

a = $\sqrt{500^{2} - 300^{2}}$

= 400

thus

$a^{2} + 300^{2}= c^{2}$

2a da/dt = 2c dc/dt

a da/dt = c dc/dt

400 (25) = 500 dc/dt

dc/dt = 1000/500

= 20 ft/s

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A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

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a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

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To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

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In this part, we are told that the area of the cell is

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While the intensity of incoming radiation (the energy received per unit area) is

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This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

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