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11111nata11111 [884]
3 years ago
11

An echo is an example of sound wave A) diffraction. B) interference. C) reflection. D) refraction.

Physics
2 answers:
DedPeter [7]3 years ago
6 0
It is an example of reflection of wave. C.
Umnica [9.8K]3 years ago
6 0

<em>The answer is c reflection because it bounces of the surfaces which makes it an echo</em>

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Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also
ozzi

Answer:

a) B_{max} = 1.784*10^{-12}

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

V_m = 160 V

a) V = V_msin(2\pi ft)

Where f = 60 Hz and V_m = 160 V

E =V/d= \frac{V_msin(2\pi ft)}{d}

For r = R

A = \pi R^2

Since \Phi_E = EA

\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}

From Ampere's Law:

\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl} where I_{encl}=0

So at r = R,

B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}

For maximum B, cos(2πft) = 1. Hence,

B_{max}=\frac{\mu_0\epsilon_0\pi fRV_m}{d}=\frac{4\pi*10^{-7}*8.85*10^{-12}*\pi*60*0.025*160}{4.7*10^{-3}}=1.784*10^{-12} T

b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:

B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}

From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:

B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}

The plot is given in the attachment.

5 0
4 years ago
What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.
Rasek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

    The tension is  T  =  440 \  N

     The length of the wire is l =  0.600 \  m

     The mass is  m  =  5.60 \ g = 0.0056 \  kg

Generally the frequency is mathematically represented as

            f  =  \frac{1}{2l}  *  \sqrt{\frac{T}{\frac{m}{l} } }

=>        f  =  \frac{1}{2 * 0.600 }  *  \sqrt{\frac{440}{\frac{0.0056}{0.600} } }

=>        f = 181 \ Hz

4 0
3 years ago
Increasing the mass attached to a spring will decrease the angular frequency of its vibrations. True or False
Scrat [10]

Answer:

decreases

Explanation:

The angular frequency of the mass is given by

\omega =\sqrt{\frac{k}{m}}

where, k is the spring constant and m be the mass of the body which is doing oscillations.

if the mass of the body increases, so the value of angular frequency decreases, as the angular frequency is inversely proportional to the square root of the mass of the body.

4 0
3 years ago
PAY ATTENTION MY QUESTION ASK FOR RADIATION!!!
Anestetic [448]

Answer:

The answer is c

Thermal energy moves within the air from the flames to the marshmallow.

Explanation:

Hope it helps

3 0
3 years ago
A train travels between two stations that are 63 km apart at an average speed of 35 m/s. How long after leaving the first statio
Semenov [28]
Distance / speed. So, 63/35. Answes is 1.8
4 0
3 years ago
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