A crate weighing 523 N rests on a plank that makes a 22.0 angle with the ground. Find the components of the crate's weight for
1 answer:
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Answer:
Length of third side = 3.97m
Explanation:
First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.
From A-B we have distance = 5.5m = Say it c
From B-C we have distance = 4.3m = Say it a
From A-C we have distance = ? = Say it b which we have to find out.
Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)
For our case it is:
b² = a² + c² - 2acCos(B) - Say it equation 1
From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.
There values are:
a = 4.3m; c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees
Now by putting the respectve values in equation 1 we have:
b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)
b² = 18.49 + 30.25 - 32.95
b² = 15.79
b = √15.79
b = 3.97m
Thus the length of third side is 3.97m.
PS: The picture of triangle is being attached for yours understanding.
This question is incomplete, the complete question;
you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
Options;
a) 200 nm; 0.9 mW
b) 100 nm, 0.0059 mW
c) 200 nm; 0 mW
d) 100 nm; 0.9 mW
e) 200 nm; 0.0059 mW
Answer:
the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Explanation:
Given that the instrument here is an interferometer.
Maximum intensity is obtained when the two waves are exactly in phase.
that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.
The phase factor of this point is taken as ∅ = 0
Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π
This corresponds to a path difference between the two waves as half of the wavelength. λ/2
The light gets reflected from the mirror.
Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)
hence, to get a path difference of λ/2 the mirror should move half of this distance only
so, the mirror should move;
= λ/4
here, wavelength is 400nm
the length moved by the mirror = 400/4 = 100 nm
The intensity is given by the equation;
=
1 +
2 + 2√
1
2cos(∅)
where
1 = 2.25 mW
2 = 2.025 mW
∅ = π
so we substitute
= 2.25 + 2.025 - 2√(2.25 × 2.025)
= 4.275 - 4.26907
= 0.0059
Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer