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salantis [7]
3 years ago
13

Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o

nly information on each sample's label is the material's coefficient of absorption. The coefficient of absorption listed on Sample A is 30%, on Sample B is 47%, and on Sample C is 62%. When your boss asks for your choice and the reasoning behind it, what would you tell him? A. Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the largest. B. Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest. C. Sample A would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the largest. D. Sample A would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest. need asp
Physics
1 answer:
garik1379 [7]3 years ago
7 0

The "coefficient of absorption" of a soundproofing material is

the fraction of sound energy that goes into the material, gets

absorbed, doesn't come out, and is never heard again.


So, sample C would be the best choice, because the percentage

of the energy in an incident wave that gets absorbed in this material

is the greatest, and the percentage of the energy that remains in a

reflected wave from this material is the smallest. (Choice-B)

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3 years ago
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An automobile accelerates 1.77 m/s2 over 6.00 s to reach freeway speed at the end of an
GaryK [48]

Answer:

Startinfg speed is 13.82 m/s

Explanation:

Use equation for realtion between start and final speed :

Vf=Vs+a*t

Vf-final speed

Vs-start speed

Vf=24.44m/s

a=1.77m/s²(acceleration)

t=6.00s(Time)

Vf=Vs+a*t

Vs=a*t-Vf

Vs=1.77m/s²*6s-24.44m/s

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6 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
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Answer:

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