Ask question

Ask question

All categories

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.

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

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

You might be interested in

The rate of spin of a disk is decreasing. A point on the disk that is near the rim of the disk will have...

NNADVOKAT [17]

At a point near the rim of the disk, it will have a<span> non-zero radial acceleration and a zero tangential acceleration. Also known as centripetal acceleration, radial acceleration takes place along the radius of the disk. On the other hand, the tangential acceleration is along the path of disk's motion.</span>

8 0

2 years ago

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

Alik [6]

Answer:

1,365 miles

Explanation:

i multiplied 455 x 3 because it traveled 455 miles an hour for three hours.

4 0

3 years ago

(show your work)

Artist 52 [7]

Answer:

I don't do physics , I'm sorry can't help you

4 0

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

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

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

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,

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

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

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
$\lambda=34.0 nm=34 \cdot 10^{-9} m$
And we can use De Broglie's relationship to find its momentum:
$p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s$

Given $p=mv$ , with m being the electron mass and v its velocity, we can find the electron's velocity:
$v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}= 2.13 \cdot 10^4 m/s$

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:
$K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J$

3 0

3 years ago