Answer:
a) d₁ = 247.8 μm
d₂ = 205.3 μm
b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm
Explanation:
a)
The formula for Michelson Interferometer is derived to be:
d = mλ/2
where,
d = distance moved
m = no. of fringes
λ = wavelength of light
For JAN, we have following data
d = d₁
m = 818
λ = 606 nm = 606 x 10⁻⁹ m
Therefore,
d₁ = (818)(606 x 10⁻⁹ m)/2
<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>
For LINDA, we have following data
d = d₂
m = 818
λ = 502 nm = 502 x 10⁻⁹ m
Therefore,
d₂ = (818)(502 x 10⁻⁹ m)/2
<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>
b)
The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,
Resultant Displacement = Δd = d₁ - d₂
Δd = 247.8 μm - 205.3 μm
<u>Δd = 42.5 μm (in the direction of JAN)</u>
A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
Answer:
Explanation:
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1 kg = 1,000 grams
To get 75kg in units of grams, we just multiply 75 by 1,000
75* 1,000 = 75,000 grams