Consider a system of two particles: ball A with a mass m is moving to the right a speed 2v and ball B with a mass 3m is moving t
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Answer:
The potential difference between points A and B is 278.95 volts.
The potential difference between points B and C is -642.10 volts.
The potential difference between points A and C is -363.15 volts.
Explanation:
Given :
Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C
Work is done to move a charge from point A to B, W₁ = 5.3 μJ
Work done to move a charge from point B to C, W₂ = -12.2 μJ
Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.
The relation between work done and potential difference is:
W = qV
V = W/q ....(1)
Using equation (1), the potential difference between points A and B is:
Substitute the suitable values in the above equation.
V₁ = 278.95 V
Using equation (1), the potential difference between points B and C is:
Substitute the suitable values in the above equation.
V₂ = -642.10 V
The potential difference between points A and C is:
V₃ = V₁ + V₂
V₃ = 278.95 - 642.10
V₃ = -363.15 V
Answer:
Explanation:
We know that the frequency of the nth harmonic is given by , where
is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:
Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):
Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):
So the tension is:
Which for our values is: