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2 years ago

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A long solenoid has a radius and length of 2.30 cm and 25 cm respectively, has 580 turns. If the current in the solenoid is incr

easing at the rate of 17.8 A/s, what is the magnitude of the induced emf produced.

1 answer:

nadya68 [22]2 years ago

3 0

Answer:

Induced emf is equal to 0.05 volt

Explanation:

We have given length of the solenoid l = 25 cm = 0.25 m

Radius of the solenoid r = 2.3 cm = 0.023 m

Area of the solenoid $A=\pi r^2=3.14\times 0.023^2=0.00166m^2$

Number of turns N = 580

Inductance of the solenoid $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7\times 580^2\times 0.00166}}{0.25}=0.0028H$

Rate of change of current $\frac{di}{dt}=17.8A/sec$

Induced emf in inductor is equal to $e=L\frac{di}{dt}=0.0028\times 17.8=0.05volt$

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V = d ÷ t --> bc d=vt
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3 years ago

what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s

yaroslaw [1]

Answer : The value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = ?

t = time = 17s

$[A_t]$ = final concentration = 0.0981 M

$[A_o]$ = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

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2 years ago

Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward

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Answer:

D = 25 miles

Explanation:

To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.

Since both bicyclists collide, we know that Xa=Xb, so:

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3 years ago

A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The

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2 years ago

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

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Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

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3 years ago