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lara31 [8.8K]
3 years ago
13

A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the mountaintop to be θ when she

is a horizontal distance x from the mountaintop. After walking a distance d closer to the mountain, she measures the angle of elevation of the mountaintop to be φ. Neglecting the height of the woman’s eyes above the ground, draw a well-labeled diagram representing this situation and find an expression for the height of the mountain, y, in terms of d, φ, and θ. Note that your expression cannot contain x.
Physics
1 answer:
valentina_108 [34]3 years ago
7 0
Let t = Theta and p = Phi
Tan t = y/x    Then x =y/Tant.
Tant = y/(x-d)  x-d = y/Tanp
y/Tant  - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp  =  d*Tanr
y(1 - Tanr/Tanp = d*Tant

y = d*Tant/(1-Tant/Tanp)
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What is the difference between a hurricane and a typhoon
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Answer:

The answer is below

Explanation:

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2 years ago
An object moves 2.5 m. This is an example of a _______. Question 3 options: direction distance velocity speed
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3 years ago
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A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction
Alja [10]

A) 1.55

The speed of light in a medium is given by:

v=\frac{c}{n}

where

c=3\cdot 10^8 m/s is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

v=1.94\cdot 10^8 m/s

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

\lambda=\frac{\lambda_0}{n}

where

\lambda is the wavelenght in quartz

\lambda_0 is the wavelength in air

n is the refractive index

For the light in this problem, we have

\lambda=355 nm\\n=1.55

Therefore, we can re-arrange the equation to find \lambda_0, the wavelength in air:

\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm

4 0
3 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

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now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0
3 years ago
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