Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object
Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula
The magnification is
Hence, The magnification is -6.05.
Answer:
Calculate the wavelength associated with an electron with energy 2000 eV.
Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J
Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
the answer is d i took the test be cause i go to a k12 onlie school
