Consider 3 polarizers. Polarizer 1 has a vertical transmission axis and polarizer 3 has a horizontal transmission axis. Taken to
gether, polarizers 1 and 3 are a pair of crossed polarizers. Polarizer 2, with a transmission axis of 300 to the vertical, is placed between polarizers 1 and 3. What is the intensity of light transmission through these polarizers given an initial intensity of light lo?
a. 0.375 lo
b. 0.25 l0
c. 0.18 lo
d. 0.09 lo
e. zero
a. 0.375 lo
b. 0.25 l0
c. 0.18 lo
d. 0.09 lo
e. zero
1 answer:
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Answer:
c)At a distance greater than r
Explanation:
For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance between the satellite and the Earth's centre
v is the speed of the satellite
Re-arranging the equation, we write
so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.
Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is
c)At a distance greater than r
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I