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Nataly [62]
3 years ago
12

Which method do acoustic use to limit reverberations in music halls

Physics
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

reverberation time appropriate to the use and size of the room, adequate balance between direct and reverberant sound, intimacy and good sound diffusion in the room to obtain a uniform sound.

Explanation:The process of ... second method measured the speed of sound propagation by the phase shift.

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The intensity of light from a star varies inversely as the square of the distance. If you lived on a planet ten times farther aw
frosja888 [35]

Answer:

the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.

That is,

Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100

Explanation:

Let the intensity of light be represented by I

Let the distance of the star be d

I ∝ (1/d²)

I = k/d²

For the earth,

Iₑ = k/dₑ²

k = Iₑdₑ²

For the other planet, let intensity be Iₒ and distance be dₒ

Iₒ = k/dₒ²

But dₒ = 10dₑ

Iₒ = k/(10dₑ)²

Iₒ = k/100dₑ²

But k = Iₑdₑ²

Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100

Iₒ = Iₑ/100

Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.

3 0
4 years ago
Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

4 0
4 years ago
30.6*X+-2.56<br> Pls help
pochemuha

Answer:  10.16

Explanation:

5 0
3 years ago
An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work
Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ

Let W' is the work must be done to lift it 290 m. It is given by :

W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ

Hence, this is the required solution.

5 0
4 years ago
B)A man walks 95 km, East, then 55 km, north. Calculate his RESULTANT
Varvara68 [4.7K]

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

Learn more about displacement at: brainly.com/question/321442

8 0
3 years ago
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