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3 years ago

The ratio of the inductive reactance to the

Physics

1 answer:

BARSIC [14]3 years ago

6 0

Answer:

The non resonance frequency of the generator is = 1201.79 Hz

Explanation:

At resonance,

f₀ = 1/2π√LC..................... Equation 1

Where f₀ = resonance frequency, L = inductance, C = capacitance

making LC the subject of the equation

LC = 1/4πf₀²..................... Equation 2

Given: f₀ = 225 Hz, and π = 3.143

Substituting these values into equation 2,

LC = 1/(4×3.143²×225²)

LC = 1/2000385.9

LC = 5×10⁻⁷

If the ratio of capacitive reactance to inductive reactance = 5.36

1/2πfC/2πfL = 5.36

1/4π²f²LC = 5.36

Where f = frequency of the non resonant

making f the subject of the equation

f = 5.36/2π√LC ............. Equation 3

Substituting the value of LC = 5×10⁻⁷ into equation 3

f = 5.36/2×3.143√(5×10⁻⁷)

f = 5.36/(6.286×0.00071)

f = 5.36/0.00446

f = 1201.79 Hz

Thus the non resonant frequency of the generator is = 1201.79 Hz

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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

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2 years ago

A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p

andreyandreev [35.5K]

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

$U_{A}+k_{A}+w_{A}=U_{B}+k_{B}$

$U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2$

$U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2$

Put the value into the formula

$U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2$

$U_{A}=88.8\ J$

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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3 years ago

What is the electromagnetic spectrum and how is it used to gather information about

Sladkaya [172]

Telescopes use lenses or mirrors to collect and focus waves from the electromagnetic spectrum, including visible light, allowing us to look at celestial objects. By studying the electromagnetic waves given off by objects such as stars, galaxies, and black holes, astronomers can better understand the universe.

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3 years ago

When was the copernican treatise published

LiRa [457]

They were published in 1542.

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4 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

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3 years ago