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jasenka [17]
3 years ago
13

The multiple reflection of a single sound wave is a/an

Physics
1 answer:
JulsSmile [24]3 years ago
6 0
<span>The multiple reflection of a single sound wave is echo</span>
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What natural phenomena could serve as alternative time standards?
KonstantinChe [14]

Answer:

Explanation:

radioactive decay, planetary orbit, speed of light, etc.

4 0
3 years ago
(a) What resonant frequency would you expect from blowing across the top of an empty soda bottle that is 18 cm deep, if you assu
V125BC [204]

Answer:

476.387 Hz

714.583 Hz

Explanation:

L = Length of tube

v = Speed of sound in air = 343 m/s

Frequency for a closed tube is given by

f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{343}{4\times 0.18}\\\Rightarrow f=476.389\ Hz

The frequency is 476.387 Hz

If it was one third full L=0.18-\dfrac{1}{3}0.18

f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{343}{4\times (0.18-\dfrac{1}{3}0.18)}\\\Rightarrow f=714.583\ Hz

The frequency is 714.583 Hz

5 0
3 years ago
What defines minerals will mark brainlest if croccted right
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3 0
3 years ago
A cylinder with a movable piston contains 2.00 gg of helium, HeHe, at room temperature. More helium was added to the cylinder an
NikAS [45]

Answer: 1.8 g

Explanation:

We start first, by calculating the amount of Helium

n = m/M

m = mass of Helium

M = molar mass if Helium

n = 2/4 = 0.5 moles

proceeding further, we use ideal gas law. PV = nRT

Then we have

P1V1/n1T1 = P2V2/n2T2

So that,

n2 = n1T1P2V2/P1V1T2

From the question, we know that, P1 = P2, and T1 = T2. So that,

n2 = n1v2/v1

n2 = (0.5 * 3.9) / 2

n2 = 1.95/2

n2 = 0.975 moles. With this, we can determine the mass, m2 of Helium

n = m/M

m = n * M

m = 0.975 * 3.9

m = 3.8

The difference between both masses are 3.8 - 2 = 1.8 g

Thus, 1.8 g of Helium was added to the cylinder

3 0
3 years ago
Read 2 more answers
A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0
3 years ago
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