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

A projectile was launched horizontally with a velocity of 468 m/s, 1.86 m above the ground. Calculate how long it would take for

elena-14-01-66 [18.8K]

Answer:

0.6 seconds

Explanation:

The time to fall from height h is ...

t = √(2h/g)

t = √(2(1.86 m)/(9.8 m/s^2)) ≈ √0.3796 s ≈ 0.616 s

It would take about 0.6 seconds for the projectile to hit the ground.

5 0

1 year ago

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In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg

harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 × $10^{-2}$ m

separation L = 0.5 mm = 0.5 × $10^{-3}$ m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 × $10^{-2}$ / 11

θ = 0.234

and we know diffraction grating for max

L sinθ = mλ

here we know m = 1 so put all value and find λ

L sinθ = mλ

0.5 × $10^{-3}$ sin(0.234) = 1 λ

λ = 2042.02 × $10^{-9}$ m

λ = 2042 nm

3 0

1 year ago

A mercury atom in the ground state absorbs 20.00 electronvolts of energy and is ionized by losing an electron. How much kinetic

sukhopar [10]

We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.

The initial energy is 20 eV = 20 J/C
The electron charge is = 1.60217662 × 10-19 coulombs

To determine the kinetic energy, we can use this equation:

KE = 20 Joules / Coulombs * 1.60217662 × 10-19 coulombs
KE = 1.25x10^20 Joules

Therefore, the amount of kinetic energy that the electron has after ionization is 1.25x10^20 Joules or 1.25x10^17 kJ.

8 0

1 year ago

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

ehidna [41]

Answer

The Value of r = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

$R_T = R_1 +R_2$

And from Ohm law

$I_s =\frac{ V}{R_T}$

$I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel is

$R_T = \frac{1}{R_1} +\frac{1}{R_2}$

$= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

=> $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

Dividing equation 2 with equation 1

=> $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

$10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3

$10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

$r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

a = 1 b = -8 c =1

$= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

$r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

$r = \ 7.87\ or \ r \ = \ 0.127$

Now r = 0.127 because it is the least value among the obtained values

4 0

1 year ago

When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.

gayaneshka [121]

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

5 0

1 year ago