Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
Sound waves travel faster through <em>solids</em> than they do through gases or liquids. <em>(C) </em>They don't travel through vacuum at all.
Example:
Speed of sound in normal air . . . around 340 m/s
Speed of sound in water . . . around 1,480 m/s
Speed of sound in iron . . . around 5,120 m/s
Answer:
False
Explanation:
When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.
As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.
ωd = ω₀√(1 - ζ)
Where ζ is called damping ratio.
For small value of ζ
ωd ≈ ω₀
[r] =6
Solve for r by simplifying both sides of the equation, then isolating the variable.
<em> </em>I hope this makes sense
