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Olenka [21]
4 years ago
10

Your amplifier is playing a 200 Hz tone. When you double the amplitude(loudness), what happens to the speed of the sound (v) and

the frequency (f)? a. v doubles, f stays the same b. v doubles, f also doubles c. v stays the same, and f doubles d. v and f both stay the same e. v and f both decrease
Chemistry
1 answer:
kherson [118]4 years ago
8 0
Its d. Speed (v) and frequency(f) both stay the same.

<span>Because Amplitude does not affect wave frequency. When you put more energy into the wave (larger amplitude) it does not affect how often you repeat a cycle. 
</span>Also speed can be altered when the medium changes through which it travels.
You might be interested in
KE and PE Math
slava [35]

Answer:

1. Answer: The bowling ball has more potential energy as it sits on top of the building. It does not have any kinetic energy because it is not moving.

2. Answer: The bowling ball has equal amounts of potential and kinetic energy half way through the fall. At the half way point, half of the potential energy has been converted to kinetic energy.

3. Answer: Just before the ball hits the ground, it has more kinetic energy. As it hits the ground the potential energy becomes zero.

4. Answer:

PE=784 J

5. Answer:  

PE = 392 J

6. Answer:

KE= 392 J

Also, since the PE and KE are equal at the half way point and PE =392 J, KE = 392 J.

7. What is the kinetic energy of the ball just before it hits the ground?

Answer:  

KE=784 J

At first I answered in the comments, but I am able to answer now. I hope this can help

6 0
4 years ago
A mass spectrum of an organic compound shows the relative abundances of M to be 40.58% and M 1 to be 7.022%. Assuming the peaks
sveticcg [70]

Answer:

The number of carbon atoms in the organic compound is 16.

Explanation:

The relative intensity of M + 1 peak (relative to M = 100) in organic compounds can be predicted by the following formula:

[M + 1] = (number of C x 1.07) ---------------------------------------------------- (1)

This formula can also be used to determine the number of carbon from the given intensity. But first, the intensity of [M + 1] relative to [M] = 100 needs to be determined, for which, consider the following calculations,

[M+1]=(\frac{M+1}{M}) (100)

[M+1]=(\frac{7.022}{40.58}) (100)

[M+1]=17.30

Using equation 1 we get,

number of C=\frac{[M+1]}{1.07}

number of C=\frac{17.3}{1.07}\\number of C = 16.168

or number of C = 16

The remaining value, 0.168, can be due to the isotopes of hydrogen and oxygen in the organic compound.

7 0
4 years ago
Which of the following has the lowest Ka?a. carbonic acid b. dihydrogen phosphate c. phosphoric acid d. bicarbonate
Naya [18.7K]

Explanation:

A compound or molecule which will have least dissociation and that is not able to given hydrogen ion easily upon dissociation will also have a low value of K_{a}.

Dissociation of the given compounds or species will be as follows.

 H_{2}CO_{3} \rightleftharpoons H^{+} + HCO^{-}_{3}

 H_{2}PO^{-}_{4} \rightleftharpoons H^{+} + HPO^{-}_{4}

 H_{3}PO_{4} \rightleftharpoons H_{+} + H_{2}PO^{-}_{4}

As, chemical formula of bicarbonate is HCO^{-}_{3} and due to the presence of a negative charge it is difficult to lose a positively charged hydrogen ion. This is because oppositely charged ions will be bonded by strong force of attraction.

Hence, it will not easily lose a hydrogen ion due to which bicarbonate has the lowest K_{a}.

Thus, we can conclude that out of the given species bicarbonate has the lowest K_{a}.    

7 0
3 years ago
HELP <br> is you increase the amount of water used to dissolve a pill, would something change?
skad [1K]

Answer:

The answer is

Explanation:

No,it is not possible.

Hope this helps....

Have a nice day!!!!

4 0
3 years ago
What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas:
riadik2000 [5.3K]

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

<em>Where the acidic dissociation constant is written as:</em>

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

<em>0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷</em>

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

<h3>pH = 3.39</h3>
5 0
3 years ago
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