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victus00 [196]
3 years ago
12

A jetliner, traveling northward, is landing with a speed of 69 m/s. Once the jet touches down, it has 750 m of runway in which t

o reduce its speed to 6.1 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing.
Physics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

Acceleration, a=-3.14\ m/s^2

Explanation:

It is given that,

Initial speed of the jetliner, u = 69 m/s

Final speed of the jetliner, v = 6.1 m/s

Distance covered, s = 750 m

We need to find the average acceleration of the jetliner. It can be calculated using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.1)^2-(69)^2}{2\times 750}          

a=-3.14\ m/s^2

So, the average acceleration of the plane during landing is -3.14\ m/s^2. Hence, this is the required solution.

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The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

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