A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency
1 answer:
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
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<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:
To maintain constant, if Q increases V also increases.
On the other hand, the potential energy can be expressed as:
In conclusion, as Q increases the potential energy also increases.