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Andru [333]
3 years ago
10

To measure the height of the cloud cover at an airport, a worker shines a spotlight upward at an angle 75° from the horizontal.

An observer D = 510 m away measures the angle of elevation to the spot of light to be 45°. Find the height h of the cloud cover. (Round your answer to the nearest meter.)
Physics
1 answer:
Yuri [45]3 years ago
4 0

Answer: The height of the cloud = 394.55 m

Explanation:

The observer is 500m away from the spotlight.

Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:

Tan 45° = h/x ... eq1

Tan 75° = h/(500- x ) ... eq2

From eq 1, Tan 45° = 1, therefore eq1 becomes:

h = x ... eq3

Put eq3 into eq2

Tan 75° = h/(500- h)

h = ( 500 - h ) Tan 75°

h = 500Tan 75° - hTan75°

h + h Tan 75° = 500 Tan 75°

h ( 1 + Tan 75° ) = 500 Tan75°

h = 500Tan75°/ (1 + Tan 75°)

h= 1866.02 / 4.73

h = 394.55m

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A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente
nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is I_3 =41.31 W/m^2

Explanation:

 According to Malu's law the intensity of a polarized light having an initial intensity I_0 is mathematically represented as

               I = I_0cos^2 \theta

Now  considering the polarizer(The polarizing disk) the equation above becomes

          I = I_0 (cos^2 \theta)^n

Where n is the number of polarizers

       Substituting  60.0W/m^2 for the initial intensity 3 for the n and 20° for the angle of rotation

           I_3 = 60 (cos^220)^3

               =41.31 W/m^2

             

     

                         

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The equation that relates two poles to the force between them and their separation is the ____ square law.
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Read 2 more answers
Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
Crazy boy [7]

Answer:

Temperature at the exit = 267.3 C

Explanation:

For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

5 0
3 years ago
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