The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>
Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:
![n_{air}=1](https://tex.z-dn.net/?f=n_%7Bair%7D%3D1)
![\theta_{liquid} = 19.38\°](https://tex.z-dn.net/?f=%5Ctheta_%7Bliquid%7D%20%3D%2019.38%5C%C2%B0)
![\theta_{air}35.09\°](https://tex.z-dn.net/?f=%5Ctheta_%7Bair%7D35.09%5C%C2%B0)
Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:
![n_1sin\theta_1 = n_2sin\theta_2](https://tex.z-dn.net/?f=n_1sin%5Ctheta_1%20%3D%20n_2sin%5Ctheta_2)
![n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}](https://tex.z-dn.net/?f=n_%7Bair%7Dsin%5Ctheta_%7Bair%7D%20%3D%20n_%7Bliquid%7Dsin%5Ctheta_%7Bliquid%7D)
![n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}](https://tex.z-dn.net/?f=n_%7Bliquid%7D%20%3D%20%5Cfrac%7Bn_%7Bair%7Dsin%5Ctheta_%7Bair%7D%7D%7Bsin%5Ctheta_%7Bliquid%7D%7D)
Replacing the values we have:
![n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}](https://tex.z-dn.net/?f=n_%7Bliquid%7D%3D%5Cfrac%7B%281%29sin%2835.09%29%7D%7Bsin%2819.38%29%7D)
![n_liquid = 1.7323](https://tex.z-dn.net/?f=n_liquid%20%3D%201.7323)
Therefore the refractive index for the liquid is 1.7323
Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min