PHYSICS COLLEGE

The record for the world’s loudest burp is 109.9 dB, measured at a distance of 2.5 m from the burper. Assuming that this sound was emitted as a spherical wave, what was the power emitted by the burper during his record burp?

Answers

Answer 1

Answer:

Answer:

Power of the source is 7.7 W

Explanation:

As we know that the sound level is measured as

$L = 10 Log (I)/(I_o)$

now we have

$L = 109.0 dB$

so we have

$109.9 = 10 Log(I)/(10^(-12))$

$I = 0.098 W/m^2$

now power of the source is given as

$P = I(4\pi r^2)$

$P = 0.098(4\pi )(2.5^2)$

$P = 7.7 W$

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20 examples of scalar quantity

Answers

Answer:

Length

Time

Mass

Temperature

Energy

Direct Current (DC)

Frequency

Volume

Speed

Amount of substance

Luminous Intensity

Density

Concentration

Refractive Index

Work

Pressure

Power

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An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is : (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line

Answers

Answer:

2.72 m

Explanation:

wavelength of sound λ = velocity / frequency

= 340 / 1200

= .2833 m

Distance of point of first constructive interference

= λ D / d ( D is distance of the screen and d is distance between source of sound.

Here D = 12.5 m

d = 1.3 m

λ D / d= ( .2833 x 12.5) / 1.3

= 2.72 m

Distance of point of first constructive interference = 2.72 m

Final answer:

The wavelength of the produced sound is approximately 0.29 m. Constructive interference occurs when the path difference between the two waves is a multiple of this wavelength, allowing you to calculate the distance between the central maximum and first maximum loud position.

Explanation:

For part (a) of the question, we need to calculate the wavelength of the sound wave. The wave speed (v) is given by the multiplication of frequency (f) and wavelength (λ). The speed of sound in air is approximately 343 m/s and given that the frequency produced by the function generator is 1200 Hz, the wavelength can be calculated using the formula λ = v / f = 343 / 1200 ≈ 0.29 m.

For part (b) the distance between the central maximum (loud) position and the first maximum along this line requires understanding of sound wave interference and constructive interference. For constructive interference to occur, the path difference between the two waves needs to be a multiple of the wavelength. Thus, in the first constructive interference position (first maximum loud position), the path difference equals one wavelength (0.29m). Since the student is walking 12.5 m away and parallel to the line between the speakers (which is the hypotenuse of a right triangle stakeout, with one side being 0.65m), we can use Pythagorean theorem to find out the distance.

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What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Answers

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^(-9)m^3$

The potential energy per unit volume is defined as the energy density.

$u = (U)/(V)$

$u= ((13.0 J))/((6.00*10^(-9) m^3))$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = (1)/(2) \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{(2u)/(\epsilon_0)}$

$E = \sqrt{(2(2.167109))/(8.85*10^(-12))}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

Final answer:

To calculate the electric field strength that would store 12.5 Joules of energy in every 6.00 mm^3 of space, we use the energy density formula. We firstly find the energy density and input it into the formula to solve for the electric field strength. The result is approximately 6.87 X 10^6 N/C.

Explanation:

The energy stored in an electric field is given by the formula U = 1/2 ε E^2. Here, U is the energy density (energy per unit volume), E is the electric field strength, and ε is the permittivity of free space.

Given that the energy stored U is 12.5 joules, and the volume is 6.00 mm^3 or 6.00X10^-9 m^3, the energy density (U) can be computed as 12.5 J/6.00X10^-9 m^3 = 2.08X10^12 Joule/meter^3.

We can solve the formula for E (electric field strength): E = sqrt ((2U)/ε). Substituting the value of ε (8.85 × 10^-12 m^-3 kg^-1 s^4 A^2), we can find E to be approximately 6.87 X 10^6 N/C.

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Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Answers

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

What are supersonic speeds

Answers

I think your answer is speed faster than the speed of sound

Answer:

speeds above 343 m/s

Explanation:

I have taken the test got 100%

A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.

Answers

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→ θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

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