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2 years ago

What can you do to prevent an echo and a reverberation in cinema halls and auditorium?

Physics

1 answer:

Gemiola [76]2 years ago

7 0

Answer:

Just make sure the distance between two objects isn't much

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A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are

zhannawk [14.2K]

Answer:

$y = 17.04 ft$

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

$y = 180 ft$

horizontal distance of the tower from mid point is given as

$x = \frac{1300}{2} = 650 ft$

now from the equation of parabola we have

$y = k x^2$

$180 = k(650^2)$

$k = 4.26 \times 10^{-4}$

now we have

$y = (4.26 \times 10^{-4})x^2$

now we need to find the height at distance of 200 ft from center

so we have

$y = (4.26 \times 10^{-4})(200^2)$

$y = 17.04 ft$

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3 years ago

What r three positive results of a variation within a population that occur due to natural selection

Phoenix [80]

Answer:

Traits, evolution, adaptive

4 0

3 years ago

If a 15g irregular shaped object is submerged in a graduated cylinder of water, the level rises from 10 ml to 25 ml, what is the

SashulF [63]

Answer:

one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/

Explanation:

6 0

3 years ago

An electron has a mass of 9.1x10-31 kg. What is

zloy xaker [14]

Answer:

3.185×10^-29 kgm/s

Explanation:

Momentum(p)=mass×velocity

=9.1×10^-31×3.5×10

=3.185×10^-29 kgm/s

4 0

3 years ago

What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh

Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled s = 9.5 m.

Substituting these values in the equation

$s= u \timest+0.5at^{2}$

$t= \sqrt{\frac{2s}{g} }$

$t=\sqrt{\frac{2\times9.5}{9.8} }$

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

4 0

3 years ago

What can you do to prevent an echo and a reverberation in cinema halls and auditorium?​

What can you do to prevent an echo and a reverberation in cinema halls and auditorium?