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3 years ago

What form of electro magnetic radiation does Wi-Fi use

Physics

1 answer:

r-ruslan [8.4K]3 years ago

4 0

Answer:

<h2>Non-Ionizing Radiation From Wireless Technology </h2>

Radiofrequency (RF) energy is a type of electromagnetic radiation. It is used to transmit signals carrying information in the form of radio waves. Radio waves are broadcast using a transmitter. When the radio waves reach a receiver, the signal is converted back into the information that was originally sent by the transmitter. When you talk on a cell phone, your cell phone acts as a transmitter and your voice is carried on radio waves to the person you are calling. Their phone acts as the receiver and converts the signal back into your voice. In addition to cell phones, other wireless devices such as radios, Wi-Fi routers, satellites, radars and pacemakers can send or receive RF energy waves.

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A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis

Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

$Y= VoT+0.5gt^{2}$

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

$Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\$

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

$V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s$

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t

Vf=0+(9.81 )(3.5)=34.335m/S

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

$V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S$

7 0

3 years ago

Which of the following is not a vector quantity?

baherus [9]

Answer: The answer is density

3 0

3 years ago

Heat energy can be transferred from a source to a receiver by I. radiation. II. convection. III. conduction. A.

nadya68 [22]

It can be transferred by I, II, and III

Hope this helps ! :}

7 0

3 years ago

Read 2 more answers

Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe

Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

$C = \frac{q}{V}$

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

$Ue = \frac{1}{2} *\frac{q^{2}}{C}$

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0

3 years ago

4) A force is applied to an object and causes and acceleration of 2.4 m/s2. The same force is

galina1969 [7]

Explanation:

Newton's second law of motion is F = ma, or force is equal to mass times acceleration

7 0

2 years ago