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lesya [120]
3 years ago
12

What phase difference between two identical traveling waves, moving in the same direction along a stretched string, results in t

he combined wave having an amplitude 1.703 times that of the common amplitude of the two combining waves? Express your answer in degrees (no need to write the unit).
Physics
1 answer:
Tom [10]3 years ago
5 0

Answer:

Explanation:

Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree

From the formula of resultant vector

(1.703I)² = I² + I² +  2 I² cosФ

2.9 I² = 2I² + 2 I² cosФ

.9I² = 2 I² cosФ

cosФ = .9 / 2

= .45

Ф = 63.25 .

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Answer:

The concentration of hydrogen ion at pH is equal to 2 := [H^+]=0.01 mol/L

The concentration of hydrogen ion at pH is equal to 6 : [H^+]'=0.000001 mol/L

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

pH=-\log [H^+]

The hydrogen ion concentration at pH is equal to 2 = [H^+]

2=-\log [H^+]\\

[H^+]=10^{-2}M= 0.01 M=0.01 mol/L

The hydrogen ion concentration at pH is equal to 6 = [H^+]

6=-\log [H^+]\\\\

[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L

Concentration of hydrogen ion at pH is equal to 2 =[H^+]=0.01 mol/L

Concentration of hydrogen ion at pH is equal to 6 = [H^+]'=0.000001 mol/L

The difference between hydrogen ion concentration at pH 2 and pH 6 :

= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L

Moles of hydrogen ion in 0.009999 mol/L solution :

=0.009999 mol/L\times 1 L=0.009999 mol

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

8 0
3 years ago
A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron
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Answer:

259.62521 seconds

Explanation:

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Here, the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s

Time = Distance / Speed

Time=\frac{38.1}{0.14675}=259.62521\ s

The time taken to reach the ship is 259.62521 seconds

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