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3 years ago

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An ac source is connected to a resistor R = 75Ω, an inductor L = 0.01 H, and a capacitor C = 4 μF. What is the phase difference

φ between the voltage of the source and the current in the circuit when the frequency of the source is equal to half the resonance frequency of the circuit?

1 answer:

zvonat [6]3 years ago

4 0

Answer:

135°.

Explanation:

R = 75 ohm, L = 0.01 H, C = 4 micro F = 4 x 10^-6 F

Frequency is equal to the half of resonant frequency.

Let f0 be the resonant frequency.

$f_{0}=\frac{1}{2\pi \sqrt{LC}}$

$f_{0}=\frac{1}{2\times 3.14 \sqrt{0.01\times 4\times 10^{-6}}}$

f0 = 796.2 Hz

f = f0 / 2 = 398.1 Hz

So, XL = 2 x 3.14 x f x L = 2 x 3.14 x 398.1 x 0.01 = 25 ohm

$X_{c}=\frac{1}{2\pi fC}$

Xc = 100 ohm

$tan\phi = \frac{X_{L}-_{X_{C}}}{R}$

tan Ф = (25 - 100) / 75 = - 1

Ф = 135°

Thus, the phase difference is 135°.

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Calculate the total number of Cl atoms in 150mL of liquid Ccl4 (d=1.589g/mL)<br>

GalinKa [24]

Answer:

The total number of Cl atoms in 150mL of liquid CCl4 is 3.73*10²⁴.

Explanation:

First you must determine the mass of CCL4 present in 150mL of CCl4. Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance, whose expression for its calculation is the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

In this case, the density value of d = 1.589 g/mL. Then, being the volume equal to 150 mL, the value of the mass can be calculated as:

mass= density*volume

mass=1.589 g/mL * 150 mL

mass= 238.35 g

Now, being the molar mass of CCl4 154 g/mol, the number of moles that 238.35 g represents is calculated as:

$moles=\frac{238.35 g}{154 \frac{g}{mol} }$

moles= 1.55

1 mole of the compound CCl4 contains 4 moles of Cl. Then, using a simple rule of three, it is possible to calculate the number of moles of Cl that 1.55 moles of CCl4 contain:

$moles of Cl=\frac{1.55 moles of CCl_{4} *4 moles of Cl}{1 mole of CCl_{4} }$

moles of Cl= 6.2

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance. In this case it can be applied as follows: if 1 mole of Cl contains 6.023*10²³ atoms, 6.2 moles of Cl how many atoms does it contain?

$atoms of Cl=\frac{6.2 moles*6.023*10^{23} atoms}{1 mole}$

atoms of Cl= 3.73*10²⁴

<u><em>The total number of Cl atoms in 150mL of liquid CCl4 is 3.73*10²⁴.</em></u>

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3 years ago

8 points)Antireflection coating can be used on the eyeglasses to reduce the reflection of light: a) A 100nm thick coating is app

NemiM [27]

Answer:

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

thickness of coating t = ?

t = λ / 4n

we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm

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3 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

or

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

or

$\eta=0.52$

or

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3 years ago

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aleksandrvk [35]

Answer:

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Explanation:

From the question we are told that

Angle of cable 2 $\theta=37.0\textdegree$

Weight of sculpture $W=5000 N$

Generally the Tension from cable 2 $T_2$ is mathematically given by

$T_2sin37\textdegree=5000N$

$T_2=5000N/sin37\textdegree$

$T_2=8308.2N$

Generally the Tension from Cable 1 $T_1$ is mathematically given by

$T_1=T_2 cos37\textdegree$

$T_1=8308.2* cos 37\textdegree$

$T_1=6655.295917 \approx 6655.3N$

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3 years ago

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Artyom0805 [142]

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3 years ago