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Brums [2.3K]
2 years ago
15

[ b) The time of reverberation of an empty hall without and with 500 audiences is 1.5 sec and 1.4 sec respectively. Find the rev

erberation time with 800 audiences. Wait​
Physics
1 answer:
Lilit [14]2 years ago
4 0

The reverberation time with 800 audiences is 0.875 seconds.

<h3>Reverberation time with 800 audience</h3>

R₁V₁ = R₂V₂

where;

  • R₁ is the reverberation time with 400 audience
  • R₂ is the reverberation time with 800 audience
  • V₁ is initial volume
  • V₂ is final volume

R₂ = R₁V₁/V₂

R₂ = (1.4 x 500) / 800

R₂ = 0.875 seconds

Thus, the reverberation time with 800 audiences is 0.875 seconds.

Learn more about reverberation time here: brainly.com/question/9278479

#SPJ1

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What do submarines translate into pictures?
prohojiy [21]
It lets the viewer know it's something to do with underwater.
6 0
3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the
laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, \Delta f = 4\ beats/s

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

\Delta f = f - f_{i}

f_{i} = f \pm \Delta f

when

f_{i} = f + \Delta f = 523 + 4 = 527 Hz

when

f_{i} = f - \Delta f = 523 - 4 = 519 Hz

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0
3 years ago
Read 2 more answers
28. Identify whether the following objects are in
Marianna [84]

Answer:

a. A baseball after it has been  hit - not in free fall

b. A rock that is thrown in the  air - not in free fall

c. The moon - free-fall

d. A paper airplane - not in free fall

e. A bird flying - not in free fall

Explanation:

  1. The free-fall is defined as the falling of an object due to the action of gravity. The object is not experiencing any other force neglecting the air resistance.
  2. If an object is in free-fall, the direction of its motion is directed towards the center of the earth. It does not have a horizontal component of velocity.
  3. If the body is under free-fall, but a centripetal force acts on it where it is equal to the gravitational force at that point. The object will have two components of velocity along the tangential line, perpendicular to the radius of the orbit.

a. A baseball after it has been  hit - not in free fall according to point 1 & 2.

b. A rock that is thrown in the  air - not in free fall according to point 1.

c. The moon - free-fall according to point 3.

d. A paper airplane - not in free fall according to point 1 & 2.

e. A bird flying - not in free fall according to point 1 & 2.

7 0
3 years ago
1 example of Deionization
xxTIMURxx [149]

Answer:

For example, the rain water does not have any salinity, except the small quantity due to the atmosphere, while the sea water has a very high salinity.

hope this helps you friend

7 0
3 years ago
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