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

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

Given:

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

3 0

2 years ago

What happens when an atom gives up loosely held valence electrons to another atom

vazorg [7]

Answer:

The number of valence electrons increases to 8 or, the atom gives up loosely held valence electrons.

Explanation:

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3 0

3 years ago

1. The picture below shows Jamal pushing with a 100-Newton (N) force on a large box. Neither

r-ruslan [8.4K]

Answer:

100 newtons

Explanation:

Given,

Jamal pushing a large box by a force, F = 100 N

Work done on the large box is, W = 0

It is because the applied force is less than the force of the friction between the two surfaces.

Yet, there will be a force that is exerted by the large box on Jamal.

According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

$F_{a} =-F_{r}$

Hence, If the action force is 100 N, then the reaction force should be in 100 N

4 0

3 years ago

Read 2 more answers

Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction

kumpel [21]

Answer:

The correct option is H

Explanation:

From the question we are told that

The index of refraction of coating is $n_1$

The index of refraction of material is $n_2$

Generally the condition for constructive for a thin film interference is mathematically represented

$2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }$

Here t represents the thickness

For minimum thickness m = 0

So

$2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }$

=> $t =\frac{\lambda}{4n_1 }$

3 0

2 years ago

A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when

Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

Throat area of the nozzle, $A^*$ = 10 cm² = 0.001 m²
The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²
Air pressure at sea level = 101.325 kPa

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

$= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896$

From supersonic isentropic table, at $\frac{A}{A^*} = 2.896$ , we can determine the following;

$M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954$

The tank pressure is calculated as follows;

$\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa$

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn more about converging-diverging nozzle design here: brainly.com/question/13889483

8 0

2 years ago