Answer:
<h2>
The magnitude of the magnetic is 0.145 T</h2>
Explanation:
Given :
Speed of proton 
Mass of proton 
Kg
The force on the proton in magnetic field is given by,
But 
(∵ Force is perpendicular to the velocity so 
)
When particle enter in magnetic field at the angle of 90° so particle moves in circle
So force is given by,
Where 
radius but in our case 0.23 m, 
C
By comparing above two equation,
T
Answer:
The frequency of the wave is 5 x 10⁹ Hz
Explanation:
Given;
wavelength of the radio wave, λ = 6.0 × 10⁻²m
radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².
Applying wave equation;
V = F λ
where;
V is the speed of the wave
F is the frequency of the wave
λ is the wavelength
Make F the subject of the formula
F = V / λ
F = (3 x 10⁸) / (6.0 × 10⁻²)
F = 5 x 10⁹ Hz
Therefore, the frequency of the wave is 5 x 10⁹ Hz
Answer: True
Explanation:
A photo detector that can respond to the entire rang of visible light can be design, it is true.
Photo detector is a device in an optical receiver which receives optical signals and convert it to electric signal. It is the key device position in front of the optical receiver.
Answer:
a) 0.167 μC/m^2
b) 1.887 * 10^4 V/m
Explanation:
Hello!
First let's find the surface charge density:
a)
Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:
σ = Q/A
Where A is the area of the satellite, which is:
A=4πr^2 = πd^2 = π(1.9m)^2
Therefore:
σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2
Now let's calculate the electric field
b)
Just outside the surface of the satellite the elctric field will be:
E = σ/ε0
Where ε0=8.85×10^−12 C/Vm
Therefore:
E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m
E = 1.887 * 10^4 V/m
Sublimation is when a solid becomes a gas. And gas to solid is deposition.
