A uniform plane wave traveling in air is incident upon a flat, lossless, and infinite in extent dielectric interface with a diel
ectric constant of 8. In the air medium, a standing wave is formed. If the normalized magnitude of the incident E-field is Eo = 1, determine the maximum value of the E-field standing wave pattern in air and the shortest distance l (in λo) from the interface where the first maximum in the E- field standing wave pattern will occur (normalized to the incident field). Select one: a. Maximum value of the E-field standing wave = 2.48; shortest distance l = λo/4 b. Maximum value of the E-field standing wave = 2.48; shortest distance l = λo/2 c. Maximum value of the E-field standing wave = 1.48; shortest distance l = λo/4 d. Maximum value of the E-field standing wave = 0.52; shortest distance l = λo/2
or one hailstone we have; Force = Mass X acceleration = 0.005kg x 9.8.} This is when the hailstone is not inclined at an angle. When the hailstone is inclined at an angle of 45, then the component of force along the glass window will be F =0.005kg x 9.8 x sin45= 0.005kg x 9.8 x 0.707= 0.0346N. Therefore, total force for the 500 hailstones would be 500x0.0346N=17.32N This force is acting on an area equal to 0.600m2 Pressure = Force per unit area = 17.32N/0.600m2 = 28.9Pa
The directions were different, so the velocities could not be the same.
However, the magnitude of the velocity (speed) was 56/2 = 28 m/s for the first car, and 84/3 = 28 m/s for the second car. These<em> average speeds are the same</em>.
