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photoshop1234 [79]

3 years ago

8

Which field in an 802.11a plcp frame are used to initialize part of the transmitter and receiver circuits?

1 answer:

defon3 years ago

4 0

What Class Is this for?

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I need the answers to these questions if u dont mind

Nonamiya [84]

there’s no photo. brainly hasent been working for me today so not sure

3 0

3 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

This implies that

$tan \theta = \frac{L}{D}$

=> $\theta = tan ^{-1} \frac{L}{D}$

$\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

$\theta =0.2865$

Substituting values into the formula for path difference

$y = 0.2 *10^{-3} sin(0.2864)$

$y = 9.997*10^{-7} \ m$

The phase difference is mathematically represented as

$\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values

$\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

$\Delta \phi =5 \pi$

Converting to degree

$\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$

the solution is subtracted by 360° in order to get the actual angle

4 0

3 years ago

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th

murzikaleks [220]

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

4 0

3 years ago

A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa

vlada-n [284]

Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second

4 0

3 years ago

You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.

Xelga [282]

<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

$\bf \: Force = Mass \times Acceleration$

<h3><u>Solution:-</u><u> </u></h3>

$\sf \: Force = Mass × Acceleration$

★ Substituting the values in the above formula,we get:

$\sf \implies \: 100 = Mass \times 5$

$\sf \implies \: Mass = \cancel\dfrac{100}{5}$

$\sf \implies \: Mass = 20 \: kg$

4 0

3 years ago

Read 2 more answers