Answer:
it increases the amplitude of the wave as it propagates.
Explanation:
Answer:
Capacitive reactance is 132.6 Ω.
Explanation:
It is given that,
Capacitance, 
Voltage source, V = 20 volt
Frequency of source, f = 60 Hz
We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :



So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.
Answer:D
Explanation:
Given
mass of object 
Distance traveled 
velocity acquired 
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy
Final Energy

where
is friction work if any


Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
Calculated weight (by experimentally) of Earth is 5.972 × 10²⁴ N
Hope this helps!
To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.
The linear mass density is given as,



The expression for the wavelength of the standing wave for the second overtone is

Replacing we have


The frequency of the sound wave is



Now the velocity of the wave would be



The expression that relates the velocity of the wave, tension on the string and linear mass density is





The tension in the string is 547N
PART B) The relation between the fundamental frequency and the
harmonic frequency is

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

Then,

Rearranging to find the fundamental frequency


