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scoray [572]
4 years ago
13

suppose you are designing a device to transmit information without wires what part of the EM spectrum will your device use and w

hy
Physics
1 answer:
miv72 [106K]4 years ago
5 0
My device will use either radio waves or light waves.
Those are the only EM waves I'm equipped to generate.

If I want to get elegant about it, I may use radio waves with
frequencies around 14.1 MHz, because those can be received
many thousands of miles away.  I know how to build an effective
antenna to radiate at that frequency, and as a licensed ham-radio
operator, I can do it legally. 
You might be interested in
A rock is thrown from a height of 2.0 m at a window that is located 9.0 m above the ground. The initial velocity of the rock is
ELEN [110]

Answer:

The value is  x = 11.81 \ m

Explanation:

From the question we are told that

   The height is h =  2.0 m

   The height of the window is d = 9.0 \ m

    The initial velocity of the rock is u =  20 \ m/s

    The angle at which it is thrown is \theta  =  40

Generally the  vertical component of the velocity of the stone is mathematically represented as

    v_y = 20 sin (40)

=>v_y = 12.86 \ m/s

Generally the height of the window from the ground is mathematically represented as using kinematic equation as

      d = h  +  v_yt + \frac{1}{2} gt^2

=>   9 = 2  +12.86 t + \frac{1}{2} * - 9.8 t^2

Here g is negative -9.8 m/s^2 because the direction of the stone is against gravity

   So

       4.9 t^2 -12.86 t + 7 =0

Solving this quadratic equation using quadratic formula we obtain

     t = 0.770 s

Generally the velocity of the stone on the x axis is mathematically represented as

       v_x =  20 * cos(40 )

=>    v_x =  15.32 \  m/s

Generally the distance between the person throwing the rock and the window is mathematically represented as

       x =  v_x * t

=>    x =  15.32 * 0.771

=>    x = 11.81 \ m

8 0
3 years ago
The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 7.17 m, measured from its tip to
gayaneshka [121]

Answer:

1.789

Explanation:

let the tangential velocity at the tip of the blade (7.17m from center) be V₁ and the tangential velocity at 4.01 be V₂.

recall that tangential velocity can be related to angular velocity, ω by the following relationship:

V = rω, where V is the tangential velocity, ω is angular velocity and r = radius.

rearranging, we get

ω = V/r

We also know that at any point in the rotation, even though the tangential velocity at 7.17m radius (V₁)   will be different from the tangential velocity at 4.01m radius  (V₂), their angular velocity will be the same, hence we can equate:

ω₁ = ω₂, or

V₁/r₁ = V₂/r₂ (rearranging)

V₁/V₂ = r₁/r₂ -----(eq 1)

We also know that centripetal acceleration can be expressed in terms of tangential velocity and radius, i.e

a =V²/r

to find the ratio of the centripetal acceleration at the tip and at r=4.01m

a₁/a₂ = (V₁²/r₁)   / ( V₂² / r₂)   (rearranging & simplify)

=  (V₁/V₂)² (r₂ / r₁)   (substituting eq1 into equation)

=  (r₁/r₂)²(r₂ / r₁)  (simplifying)

= (r₁/r₂)    (substituting r₁=7.17 and r₂=4.01)

= 7.17 / 4.01

= 1.789

7 0
4 years ago
A magnetic field applies forces on:<br>a)static charges<br>b)moving charges<br>c)water flow​
lisabon 2012 [21]

Answer:

moving charges b)

I think this is the answer

4 0
4 years ago
Read 2 more answers
Labels on chemical containers do not include all the same information as found on the SDS true or false
raketka [301]

the answer is: True.

3 0
4 years ago
Read 2 more answers
If the plant produces electric energy at the rate of 1. 5 gw , how much exhaust heat is discharged per hour?
expeople1 [14]

The amount of exhaust heat discharged per hour from the plant which produces electric energy at the rate of 1. 5 gw, is 9.36×10¹⁹ J/h.

<h3>How to calculate the heat discharged per hour?</h3>

The heat is discharged per hour is equal to the different of total power and actual power.

The plant produces electric energy at the rate of 1. 5 gw. Let assume the efficiency of this power plant is 64% between the temperature of 660 degree C to 330 degree Celsius.

Thus, the maximum efficiency is,

\eta_{max}=1-\dfrac{660+273}{330+273}\\\eta_{max}=0.55

The total power is,

P_t=\dfrac{P_A}{\eta_{max}\times0.64}\\P_t=\dfrac{1.5}{0.55\times0.64}\\P_t=4.26\rm\; GW

Thus, the heat discharged per hour is,

Q=4.26-1.5\\Q=2.76\text{GW}\\

Multiply the value with 3600 to convert it in s/h,

Q=2.76\times10^9\times3600\text{ J/h}\\Q=9.36\times10^{19}\text{ J/h}\\

Thus, the amount of exhaust heat discharged per hour from the plant which produces electric energy at the rate of 1. 5 gw, is 9.36×10¹⁹ J/h.

Learn more about the heat discharged per hour here;

brainly.com/question/19666326

#SPJ4

6 0
2 years ago
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