Physics: The signal from the oscillating electrode is fed into an amplifier, which reports the measured voltage as an rms value, 1.0 nV . What is the potential difference between the two extremes

The signal from the oscillating electrode is fed - id10076481, sssssaaaaaddddd7485

rachel63892

14.02.2022

The value is $V = 2.8284 *10^{-9 } \ Volts$

Explanation:

From the question we are told that

The measure voltage is $E_{rms} = 1.0 \ n V = 1.0 *10^{-9} \ V$

Generally the peak voltage is mathematically represented as

$E_{max} = \sqrt{2} * E_{rms}$

=> $E_{max} = \sqrt{2} * 1.0 *10^{-9}$

=> $E_{max} = 1.4142 *10^{-9} \ volts$

Generally the potential difference between the two extremes is mathematically represented as

$V = 2 * E_{max}$

=> $V = 2 * 1.4142 *10^{-9}$

=> $V = 2.8284 *10^{-9 } \ Volts$

0,0(0 оценок)

axxpjr12

30.10.2021

Step-by-step explanation:

g(r)=25-3r

subsitute r for 4

g(4)=25- (3*4)

g(4)=25-12

g(4)=13

0,0(0 оценок)