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taurus [48]
3 years ago
14

What instrument can be used to show the shape of a sound wave?

Physics
1 answer:
mafiozo [28]3 years ago
7 0
"Oscilloscope"  can be used to show the shape of a sound wave

Hope this helps!
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Directions: Select the choice that best fits each statement. The following question(s) refer to the following concepts related t
borishaifa [10]

Answer:

(E) Second law of thermodynamics

Explanation:

The second law of thermodynamics can be understood according to Clausius' words: In an isolated system, no process can occur if a decrease in the total entropy of the system is associated with it. These processes are associated with energy transformations, in which a variable is introduced, called entropy that indicates the notion of disorder. Therefore, in any isolated process, the disorder can only grow.

4 0
3 years ago
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What is the force on a 15.5 kg ball that is falling freely due to the pull of gravity
velikii [3]

Answer:

151.9 N

Explanation:

Force = mass x acceleration

Acceleration due to gravity is 9.8 m/s^2 (you should memorize this number).

F = ma

F = (15.5)(9.8)

F = 151.9

3 0
3 years ago
1. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
Kobotan [32]
 0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

7.2 - 5 = 2.2 kg m/s is the motmentum of the ball now 

 the velocity  is 2.2/0.4 = 5.5 m/s
7 0
3 years ago
Reduce<br> Reflects<br> Transfers<br> Stops
Bond [772]
I think transfers is the answer
4 0
2 years ago
A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0
3 years ago
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