(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second
<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>
The 2 applications of electrostatic induction are:
1-Photocopier
2-Spray Painting
Answer:
a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m
b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m
Explanation:
(a) Radiowave wavelength= λ = c/f
As we know, Radiowave speed in the air = c = 3 x 10^8 m/s
f = frequency = 604 kHz = 604 x 10^3 Hz
Hence, wavelength = (3x10^8/604x10^3) m
λ
= 496.69 m
So the height of the antenna BROADCASTING AT 604 kHz = λ /4 = (496.69/4) m
= 124.17 m
(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz (1kHZ = 1000 Hz)
Hence, wavelength = λ = (3 x 10^8/1710 x 10^3) m
λ= 175.44 m
So, height of the antenna = λ /4 = (175.44/4) m
= 43.86 m
To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.
From the law of Malus intensity can be defined as
Where
Angle From vertical of the axis of the polarizing filter
Intensity of the unpolarized light
The expression for the intensity of the light after passing through the first filter is given by
Replacing we have that
Re-arrange the equation,
Re-arrange to find \theta
The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°
