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Lelu [443]
3 years ago
15

When making a sound source near a reflective surface, such as making a person at a podium, the sound can reflect off of the surf

ace an create phase interference known as comb filtering. An approach to solve this problem is...
Physics
1 answer:
Liula [17]3 years ago
5 0

Answer:

1. By removing the reflector or reflecting elements

2. By using absorbing material on the reflecting surfaces.

Explanation:

To solve problems caused by reflecting surfaces, you can remove the reflector or reflecting elements. However, sometimes it is enough to turn/move the reflecting element and change the reflections. Furthermore, using absorbing material on the reflecting surface can help reduce comb filtering as well.

In some cases, it is possible to place the sound source so close to the reflecting surface, that there is no suspension between the direct sound and the reflected sound.

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An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part
Varvara68 [4.7K]

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

v=\frac{Bqr}{m}

v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

T=\frac{2\pi r}{v}

T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}

T = 1.09 x 10^-7 s

(c) The formula for the kinetic energy is

K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}

K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

V = \frac{K}{q}

V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}

V = 70312.5 V

5 0
3 years ago
A lightning bolt occurs when billions of protons are transferred at the same time. ____________________
RSB [31]
True. You're suppose to answer true or false, right?
3 0
3 years ago
An LC circuit consists of a 3.14 mH inductor and a 5.08 µF capacitor. (a) Find its impedance at 55.7 Hz. 563.57 Correct: Your an
ANEK [815]

Answer:

a)

z=561.7

b)

z=214.1

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

a)

f = frequency = 55.7 Hz

Impedance is given as

z=\frac{1}{2\pi fC} - 2\pi fL

z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)

z=561.7

b)

f = frequency = 11000 Hz

Impedance is given as

z= - \frac{1}{2\pi fC} + 2\pi fL

z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)

z=214.1

4 0
3 years ago
4. In which of the following circuits will the bulb glow?
Zanzabum

Answer:

I believe is C.

Explanation:

its quite obvious:)

4 0
3 years ago
Two charges of +4x10-SC and -4x10-5C are placed 1 meter apart. Using Coulomb's law, the force between them is 14.4 N.
Snowcat [4.5K]

Answer:

d. the force is reduced by one quarter to 3.6 n

Explanation:

6 0
2 years ago
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