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Lesechka [4]
3 years ago
14

Why does sound wave bend around obstacle more than light waves​

Physics
1 answer:
hjlf3 years ago
4 0

Answer:

Since the wavelength of visible light is on the order of 0.5 microns, or 0.0005 mm, light will only diffract when going through very narrow openings. Sound waves, on the other hand, have a wavelength on the order of 1 meter and diffract very easily. This allows sound waves to bend around corners.

Explanation:

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if you transfer equal amounts of geat to 1kg of water and 1kg of cooper, which will change tempature more
Angelina_Jolie [31]

If you transfer equal amounts of geat to 1kg of water and 1kg of cooper, the temperature of the cooper will change more . . . it'll get more degrees warmer than the water will.

That's because the specific geat of water is greater than the specific geat of cooper.

(This just means it takes more geat to warm some mass of water by some amount than it takes to warm the same mass of cooper by the same amount.)  

4 0
3 years ago
Muatan listrik yang mengalir dari potensial tinggi ke potensial rendah disebut
igomit [66]

Answer:

Muatan listrik yang mengalir dari potensial tinggi ke potensial rendah disebut

elektron

Explanation:

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8 0
3 years ago
The material through which a mechanical wave travels is a.a medium.b.empty space.c.ether.d.air
Elza [17]
<h2>Answer: Medium </h2>

The medium is the main factor that differentiates a mechanical wave from an electromagnetic wave, since the first can not propagate without its existence, while the second can propagate regardless of whether the medium exists or not.

In addition, it is the medium that will define, the propagation speed of the wave, according to its specific physical characteristics.

Therefore, the <u>correct answer</u> is a.

5 0
3 years ago
Need help asap pls
ozzi
Detailed Explanation:

1) Rusting of Iron

4Fe + 3O2 + 2H2O -> 2Fe2O32H2O

Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4

Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4

2) Fermentation of sucrose…

C12H22O11 + H2O -> 4C2H5OH + 4CO2

Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12

Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12

Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.

All the Best!
8 0
3 years ago
Read 2 more answers
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

3 0
3 years ago
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