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DanielleElmas [232]
3 years ago
11

In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the

inductor is I/2, the energy stored in the capacitor is
Physics
1 answer:
Deffense [45]3 years ago
5 0

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  U =\frac{Qmax^2}{C}

So conclude C doesn't change substantially as well as,

When,

⇒  Q=\frac{Qmax}{2}

⇒  Q^2=\frac{Qmax^2}{4}

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

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A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

4 0
3 years ago
If it takes 50m to stop a car initially moving at 25mps, what distance is required to stop a car moving 50mps?
Anastaziya [24]
100m... Hope this helps :)

5 0
3 years ago
Read 2 more answers
45 grams in kilo grams
Hunter-Best [27]

Answer:

0.045 kg

Explanation:

1000 grams = 1kg

7 0
3 years ago
Read 2 more answers
A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.
Simora [160]

Answer:

\mu_k=\frac{a}{g}

Explanation:

The force of kinetic friction on the block is defined as:

F_k=\mu_kN

Where \mu_k is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

N=mg\\F_k=F=ma

Replacing this in the first equation and solving for \mu_k:

ma=\mu_k(mg)\\\mu_k=\frac{a}{g}

6 0
3 years ago
Three points A, B, and C of unknown charges are at the corners of an equilateral triangle.
nasty-shy [4]
Answer: option d. 

Q_A= \frac{1}{2} Q_B=- \frac{1}{3} Q_C

Explanation:


1) The direction of the field lines inform about the sign of the charges.

The field lines <span>extend from the positive charges to the negative charges, so you can conclude that the charge C is positve and both charge A and charge B are negative:
</span><span>
</span><span>
</span><span>Charge C: positive
</span><span>
</span><span>Charge A: negative
</span><span>
</span><span>Charge B: netative
</span>

2) The density of the lines (number of lines in a region) inform about the magnitude of the electric field.

Since the charges are at the same distance, the magnitude of the electric field informs directly about the magnitude of the force and that about the magnitude of the charges.

Since, there are the double of lines between C and B than between C and A, the magnitude of charge B is the double than the magnitud of charge A.

From the five options given (a throug e) the only that is consistent with that charges A and B have the same sign, that charge C has different sign, and that charge B is the double of charge A is:

Q_A= \frac{1}{2} Q_B=- \frac{1}{3} Q_C
3 0
3 years ago
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