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WARRIOR [948]
1 year ago
11

In aviation, it is helpful for pilots to know the cloud ceiling, which is the distance between the ground and lowest cloud. The

simplest way to measure this is by using a spotlight to shine a beam of light up at the clouds and measuring the angle between the ground and where the beam hits the clouds. If the spotlight on the ground is 0.75 km from the hangar door. What is the cloud ceiling?
Physics
1 answer:
saw5 [17]1 year ago
4 0

Answer:

b = a \frac{cos(\alpha)}{sin(\alpha)} = a Cot(\alpha)

Explanation:

In order to calculate the height reached by the beam when it hit the cloud, it is possible to use the following mathematical equations based on the Pythagoras' theorem:

b  = \sqrt[2]{H^{2} - a^{2}}

b = \sqrt[2]{H^{2} - H^{2} sin^{2} (\alpha)} \\b = H \sqrt[2]{1 - sin^{2} (\alpha)}

And if:

H = \frac{a}{sin(\alpha)}

Then:

b = \frac{a}{sin(\alpha)} \sqrt{1 - sin^{2}(\alpha)}\\b = \frac{a}{sin(\alpha)} \sqrt{1 - 1 + cos^{2}(\alpha)}\\

b = \frac{a}{sin(\alpha)} \sqrt{cos^{2}(\alpha)}\\b = a \frac{cos(\alpha)}{sin(\alpha)} = a Cot(\alpha)

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The speed of sound in dry air at 20 °C is 343.5 m s-1, and the frequency of the sound from the note C# above middle C on the pia
Nastasia [14]

Answer:

1.23917 m

0.14323 s

Explanation:

v = Speed of sound in dry air at 20 °C = 343.5 m/s

f = Frequency of note C# = 277.2 /s = 277.2 Hz

λ = Wavelength

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343.5}{277.2}\\\Rightarrow \lambda=1.23917\ m

Wavelength = 1.23917 m

Distance the wave needs to travel is 49.2 m

Time = Distance / Speed

\text{Time}=\frac{49.2}{343.5}=0.14323\ s

Time taken for the sound to travel across the concert hall is 0.14323 s

4 0
2 years ago
Two objects exert a gravitational force on 8 N on one another. What would that force be if the mass of BOTH objects were doubled
seropon [69]
<span>Based on Newton's law of universal gravitation, the equation for the gravitational force exerted by an object on another object is given by:
F = Gm1m2/(r^2)
where G is the universal gravitational constant, F is the gravitational force exerted, m1 is the mass of the first object, m2 is the mass of the second object, and r is the separation distance between the two objects.
If the mass of both objects were doubled, then we would have: m1' * m2' = (2m1) * (2m2) = 4m1m2. Assuming r stays constant (G is a constant so that won't change anyway), then this means that the new force will be 4 times greater, ie 8N * 4 = 32N of gravitational force. </span>
4 0
2 years ago
Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is ca
Umnica [9.8K]

Answer:

right A, B, C, D

Explanation:

They ask which statements are true

A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping

B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation

C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.

D) Right. It is a resonant process the driving force keeps the oscillation of the system

7 0
2 years ago
You have found a treasure map that directs you to start at a hollow tree, walk 300 meters directly north, turn and walk 500 mete
Dmitry_Shevchenko [17]

Answer:633 m

Explanation:

First we have moved 300 m in North

let say it as point a and its vector is 300\hat{j}

after that we have moved 500 m northeast

let say it as point b

therefore position of b with respect to a is

r_{ba}=500cos(45)\hat{i}+500sin(45)\hat{j}

Therefore position of b w.r.t to origin is

r_b=r_a+r_{ba}

r_b=300\hat{j}+500cos(45)\hat{i}+500sin(45)\hat{j}

r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}

after this we moved 400 m 60^{\circ} south of east i.e. 60^{\circ} below from positive x axis

let say it as c

r_{cb}=400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=r_{b}+r_{cb}

r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}

magnitude is \sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}

=633.052

for directiontan\theta =\frac{250\sqrt{2}+300-200\sqrt{3}}{250\sqrt{2}+200}

tan\theta =\frac{307.139}{553.553}

\theta =29.021^{\circ} with x -axis

7 0
1 year ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
1 year ago
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