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snow_tiger [21]
3 years ago
12

A plane rises from​ take-off and flies at an angle of 5 degrees5° with the horizontal runway. When it has gained 800800 ​feet, f

ind the​ distance, to the nearest​ foot, the plane has flown.
Physics
1 answer:
dedylja [7]3 years ago
4 0

Answer:

distance=9188149.567feet

Explanation:

Given Data

Angle α=5°

height h=800800 feet

To find

Distance r

Solution  

As we Know that

Sin\alpha =(\frac{Perpendicular}{hypotenuse} )\\Sin\alpha =\frac{h}{r}\\ r=\frac{h}{Sin\alpha}\\ r=\frac{800800feet}{Sin(5^{o} )}\\ r=9188149.567feet

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How many miles did a plane travel if it flew 455 miles per hour in 3 hours? What formula did you have to use?
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Answer:

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3 years ago
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3 years ago
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips
Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

f = \frac{v}{\lambda}

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\frac{f'}{f} = \frac{v'}{v}

The beat frequency heard when the two strings are sounded simultaneously is

f_{beat} = f-f'

f_{beat} = f(1-\frac{f'}{f})

f_{beat} = f(1-\frac{v'}{v})

We have the velocity of the transverse waves in stretched string as

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{200N}{\mu}}

And,

v' = \sqrt{\frac{196N}{\mu}}

Therefore the relation between the two is,

\frac{v'}{v} = \sqrt{\frac{192}{200}}

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4 0
2 years ago
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
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Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
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This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
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3 0
3 years ago
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