Answer:
<h2><em>
3.067MHz</em></h2>
Explanation:
The formula for calculating the voltage across an inductor is expressed as

Given parameters
current amplitude I = 1.50mA = 1.5*10⁻³A
inductance L = 0.450mH = 0.450*10⁻³H
Voltage across the inductor
= 13.0V
Required
frequency f
Substituting the given parametres into the formula, we have;

<em>Hence, the frequency required is 3.067MHz</em>
<h3>Longitudinal Waves:</h3>
- Longitudinal waves are waves in which particles moves in the direction parallel to the direction of wave.
- Longitudinal wave are one directional waves.
- Sound waves and tension force applied on spring are examples of longitudinal waves.
<h3>Transverse Waves:</h3>
- Transverse waves in which particles move perpendicular to the direction of wave propagation.
- Transverse wave are two directional waves.
- Ripples formed on water and all electromagnetic waves are examples of transverse waves.
Answer:
Elevation =31.85[m]
Explanation:
We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.
We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.
The point where the ball is thrown shall be taken as a reference point of potential energy.
![E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D)
m = mass of the ball = 300 [gr] = 0.3 [kg]
v = initial velocity = 25 [m/s]
![E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20m%2A%20v%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%200.3%2A%20%2825%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D%2093.75%20%5BJ%5D)
![93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m]](https://tex.z-dn.net/?f=93.75%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%5Bm%5D%5C%5Creplacing%5C%5Ch%3D%5Cfrac%7BE_%7Bk%7D%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B93.75%7D%7B.3%2A9.81%7D%20%5C%5Ch%3D31.85%5Bm%5D)
Answer:
Since v = (x(2) - x(1)) / t
point 2 obviously has the greatest displacement in a given time
Also, point 2 is the steepest line on this graph.