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Westkost [7]
3 years ago
5

How do odd-shaped ceilings, decorative panels, draperies, and glass windows affect echo and noise?

Physics
1 answer:
vesna_86 [32]3 years ago
6 0
I think they decrease echo and reduce noise, they do this by either absorbing vibrations or by scattering the sound so that echoes arrive at different times rather than reverberating as a standing wave. An echo is a reflection of a sound that arrives at the listener with a delay after the direct sound. The delay is usually proportional to the distance of the reflecting surface from the source and the listener.
You might be interested in
Describe the relationship between the length and period of a pendulum in the language of direct proportions
Wittaler [7]

The period of the pendulum is directly proportional to the square root of the length of the pendulum

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L, T\propto \sqrt{L}. This means that

The period of the pendulum is directly proportional to the square root of the length of the pendulum

From the equation, we also notice that the period of a pendulum does not depend on its mass.

#LearnwithBrainly

3 0
3 years ago
Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend
Rasek [7]

Answer:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

3 0
4 years ago
A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.30 m/s . The coefIficient o
Artemon [7]

Answer:

Explanation:

Mass =11.2kg

Constant velocity =3.3m/s

μk=0.25

Since the body is moving in constant velocity, then the acceleration is zero(0).

ΣF = Σ(ma)

The normal force acting on the body is upward and the weight is acting downward

Then ΣFy=0

Therefore, N=W

W=mg=11.2×9.8=109.76N

So, N=W=109.76N

Frictional force is given as

Fr=μkN

Fr=0.25×109.76

Fr=27.44N

Frictional force acting against the motion is 27.44N

Then the forward force moving the body forward

ΣF = Σ(ma)

Since a = 0

Then,

ΣF = 0

F-Fr=0

Then F=Fr

So the force moving the body forward is 27.44N

8 0
3 years ago
In comparing molar specific heat for gases under constant pressure CP and constant volume CV, we conclude that (more than one co
Sauron [17]

Answer:

b. Specific heat increases as the number of atoms per molecule increases.

c. Specific heat at constant pressure is higher than at constant volume.

d. Monatomic gases behave like ideal gases.

Explanation:

Specific heat of the gas at constant pressure is usually higher than that of the volume.

i.e.

Cp - Cv = R

where R is usually the gas constant.

However, monoatomic gases are gases that exhibit the behavior of ideal gases. This is due to the attribute of the intermolecular forces which plays a negligible role. Nonetheless, the case is not always true for all temperatures and pressure.

Similarly, the increase in the number of atoms per molecule usually brings about an increase in specific heat. This effect is true as a result of an increase in the total number associated with the degree of freedom from which energy can be separated.

Thus, from above explanation:

Option b,c,d are correct while option (a) is incorrect.

4 0
3 years ago
A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci
Anettt [7]

Answer:

The radius is  r = 3.1905 \ m

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  d =  2.70 \ m

        The refractive index of the liquid is  n_i  =  1.31

Now the critical value is mathematically represented as

         \theta =  sin ^{-1} [\frac{1}{n_i} ]

substituting values

         \theta =  sin ^{-1} [\frac{1}{131} ]

         \theta  =  49.76^o

Using SOHCAHTOA rule we have that

         tan \theta =  \frac{ r}{d}

=>     r =  d * tan \theta

substituting values  

        r =  2.7 * tan (49.76)

        r = 3.1905 \ m

         

5 0
4 years ago
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