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adell [148]
2 years ago
9

Which is not a device for reproducing sound?

Physics
1 answer:
Alona [7]2 years ago
7 0
Phonograph. Hope that helps
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What wave have a frequency of less then 20hz
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Answer: Infrasound

Explanation: Sounds with frequencies below 20 hertz are called infrasound. Infrasound is too low-pitched for humans to hear. Sounds with frequencies above 20,000 hertz are called ultrasound

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when you hold a warm bowl of soup the temperature of the palms of your hands increases this is an example of
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3 years ago
Thomas is climbing Mt. Everest. What happens as he climbs farther up the mountain?
lakkis [162]

Answer:

B. the air pressure decreases

Explanation:

As elevation increases, there is less overlying atmospheric pressure mass, so that atmospheric pressure decreases with increasing elevaton.

4 0
3 years ago
Read 2 more answers
A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/
Furkat [3]

Answer:

1) 23.45 rad/s²

2) 2.7 m/s²

3) t= 1.6 s

4) x ≈ 11 m

5) vfinal = 4.45 m/s

6) KErot = 16.2 J

    KEtran = 41 J

    KErot < KEtran

Explanation:

Step 1: Data given

mass bowling ball = 4.1 kg

radius = 0.117 meter

initial speed = 8.9 m/s

1) What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?

α = a / r = 2.774 m/s² / 0.117m = 23.45 rad/s²

2)What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?

a = µ*g = 0.28 * 9.8m/s² = 2.744 m/s² ≈ 2.7 m/s²

3) How long does it take the bowling ball to begin rolling without slipping?

This begins when ω = v / r

with

⇒ ω = α*t = 23.45 rad/s² * t

⇒ v = Vo - a*t = 8.9m/s - 2.744m/s²*t

This gives us:

23.45rad/s² * t = (8.9m/s - 2.744m/s²*t) / 0.11m

2.744*t = 8.9 - 2.744*t

t = 8.9 / 5.488 = 1.622 s ≈ 1.6 s

4) How far does the bowling ball slide before it begins to roll without slipping?

x = Vo*t - ½at² = (8.9*1.622 - ½*2.744*(1.622)²) m = 10.82 m ≈ 11 m

5) What is the magnitude of the final velocity?

v = Vo - at = 8.9m/s - 2.744m/s² * 1.622s = 4.45 m/s

6) After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:

trans KE = ½ * 4.1kg * (4.45m/s)² =40.595 J ≈ 41 J

I = (2/5)mr² = (2/5) * 4.1kg * (0.117m)² = 0.0224 kg·m²

ω = v/r = 4.45m/s / 0.117m = 38.03 rad/s, so

rot KE = ½Iω² = ½ * 0.0224kg·m² * (38.03rad/s)² = 16.2 J

16.2 J < 41 J

KErot < KEtran

(For a rolling solid sphere, KErot ≈ 2/5 * KEtran)

6 0
3 years ago
Please guys i need help in this
JulijaS [17]

Answer:

10 seconds

Explanation:

We have the equation V = at  (speed = acceleration x time)

We want to find the time, so can rearrange to T = V/a (time = speed / acceleration).

From the question, we know V is 5 and a is 0.5.

Now we can substitute that into our equation: 5/0.5 = 10.

So the time is 10 seconds.

Hope this helps! Let me know if you have any questions :)

8 0
2 years ago
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