All categories

3 years ago

Explain how the absorption of sound waves differs from the transmission of sound waves

1 answer:

4 0

When a sound wave meets an obstacle, some of the sound is reflected back from the front surface and some of the sound passes into the obstacle material, where it is absorbed or transmitted through the material.

Reflection and absorption are dependent on the wavelength of the sound. The percentage of the sound transmitted through an obstacle depends on how much sound is reflected and how much is absorbed. We are assuming that the obstacle is relatively large, such that no sound passes around the edges.

You might be interested in

If the earth can travel 1800 km in 1 minute what is it’s speed in km/s

Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

4 0

3 years ago

B. How can you tell where sugar enters the blood?

Korolek [52]

Answer:

Sugar can’t enter cells directly

Explanation:

So when blood sugar level rises, cells in the pancreas signal for the release of insulin into the bloodstream. The insulin attaches to the sugars and signals cells to let it enter with the attached sugar. Insulin is known as the key that unlocks cells.

I hope I helped

8 0

3 years ago

Given three different locations on Earth's surface, where will the weight of a person be greatest?

Feliz [49]

Answer:

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that

$F = \frac{Gm_{1} m_{2} }{r^{2} }$

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of $G, m_{1} ,m_{2}$ , we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

4 0

2 years ago

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel

nexus9112 [7]

Answer:

The electric flux is $280\ \rm N.m^2/C$

Explanation:

Given:

Radius of the disc R=0.50 m
Angle made by disk with the horizontal $\theta=30^\circ$
Magnitude of the electric Field $E=713.0\ \rm N/C$

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

$\phi=\int E.dA$

where

$\phi$ is the total Electric Flux
E is the Electric Field
dA is the Area through which the electric flux is to be calculated.

Now according to question we have

$=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C$

Hence the electric flux is calculated.

8 0

3 years ago

Can someone help me ASAP

Phantasy [73]

Well I don't know. Let's actually LOOK at the picture and see if that helps.

A, B, C, and D all have the same TOTAL length, but A has the most waves crammed into that same total length.

By golly, that means the length of each wave in A must be shorter than each wave in B, C, or D.

The correct choice is A . Looking at the picture did the trick !

7 0

2 years ago