Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Answer:
r1 -r2 = 3.75cm
Explanation:
Check the attached file for the solution
The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
<h3>What is an electric field?</h3>An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
The given data in the problem is given by;
E is the electric field = (200 N/C)
d is the distance = 5.0 cm.=0.05 m
Q is the charge of electrons= 1.602 x 10^-19 C
The formula for electric potential is given by;
The work is defined as the product of the potential difference and charge of an electron.
Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
To learn more about the electric field refer to the link;