The amount of friction depends on the force pushing the surfaces together. If this force increases, the hills and valleys of the surfaces can come into closer contact. The close contact increases the friction between the surfaces.
Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
Astronomers use the doppler effect to study the motion of objects across the Universe.
Given: Wavelength λ = 410 nm convert to Meters m = 4.10 x 10⁻⁷ m
Speed of light c = 3 x 10⁸ m/s
Required: Frequency f = ?
Formula: c = λf
f = c/λ
f = 3 x 10⁸ m/s/4.10 x 10⁻⁷ m
f = 7.32 x 10¹⁴/s or 732 Thz (Terahertz)
