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Luba_88 [7]
3 years ago
15

Based on these answers how can I determine when electric field strength is affected or not

Physics
1 answer:
Stolb23 [73]3 years ago
6 0

It's not that hard. Let me explain...

There are two ways to remember this.

The first one is by using the formula for the electric strength (E)

E = k × (Q / r^2) , where:

  • k is a constant
  • Q is the "central" charge. The one producing the electric field. <u>Not</u> the one we use to test it.
  • r is the distance between the charges

As you can see, field's strength is proportional to the magnitude of Q, and inverse proportional to the square of the distance. The field does not care how much you alter the charge you positioned in ot to test it.

Although that is the only way to work out the problem, there are many ways to remember it. Given that i have difficulty in remembering formulas, i always try to derive it from something i know.

By definition: E is equal to the force acted on the test charge, divided by its value.

E = F / q

  • F --> Coulomb's force
  • q --> test charge

Since i remember that:

F = k ( Q × q) / r^2

just take out from the expression the q term, since you divided by it, and all it remains is the strength of the field.

In other words, the strength changes as long as something from the expression is altered.

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NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a
jeyben [28]

Answer:

λ₁ = 2.50 10⁻² m,   λ₂ = 1.66 10⁻² m  

Explanation:

Microwave communication is very efficient because it does not have atmospheric interference, for which it is widely used and has been regulated to avoid interference, the ku band is in the range between 12 and 18 GHz.

Let's calculate the wavelength for the two extreme frequencies of this band

wavelength and frequency are related

         c = λ f

          λ = c / f

f₁ = 12 GHz = 12 10⁹ Hz

          λ₁ = 3 10⁸ /12 10⁹

           λ₁ = 2.50 10⁻² m

f₂ = 18 GHz = 18 10⁹ Hz

          λ₂ = 3 10⁸ /18 10⁹

          λ₂ = 1.66 10⁻² m

Unfortunately in your exercise the specific frequency is not fired, for significant figures they must be the same number as the figures of the frequency, in general the frequency has 3 or 4 significant figures

8 0
3 years ago
At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw
allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

f_i = f_0 \frac{c}{c+v}

Where

f_h= Approaching velocities

f_i= Receding velocities

c = Speed of sound

v = Emitter speed

And

f_h = f_0 \frac{c}{c+v}

Therefore using the values given we can find the velocity through,

\frac{f_h}{f_0}=\frac{c-v}{c+v}

v = c(\frac{f_h-f_i}{f_h+f_i})

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

v = 353(\frac{2.4-1}{2.4+1})

v = 145.35m/s

Therefore the cars goes to 145.3m/s

7 0
3 years ago
The curved movement of air or water is the result of which of these?
ella [17]
<h3><u>Answer;</u></h3>

B. the rotation of Earth on its axis

<h3><u>Explanation;</u></h3>
  • <em><u>The Coriolis effect describes how the Earth's rotation steers winds and surface ocean currents. The Coriolis effect causes the motion of a freely moving object to appear as a curve.</u></em>
  • <em><u>Therefore, since air and water move freely, the Coriolis effect makes their movement to be curved. This is why the movement of winds and oceans  does not follow a straight line but it is bent and curved.</u></em>
8 0
4 years ago
Read 2 more answers
The blades of a metal cutter are shorter than the blades of paper cutter scissor. Give reason​
Sladkaya [172]

Answer:

Metal is a tougher material to cut though, so the blades must be shorter to create more pressure to break through the metal. Paper on the other hand is easier to cut through so the blades can be longer in order to cut more in each snip.

8 0
3 years ago
Attempting to impress the skeptical patrol officer with your physics knowledge, you claim that you were traveling so fast that t
horsena [70]

Answer:

v_r = 1.268 × 10⁸ mi/hr

Explanation:

We are given;

wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m

wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m

Frequency is given by the formula;

f = v/λ

Where v is speed of light = 3 x 10^(8) m

Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz

Similarly, Frequency of yellow light;

f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz

To find the speed of the car, we will use the formula;

f = f_o[(c + v_r)/c)]

Where c is speed of light and v_r is speed of car.

Making v_r the subject;

cf/f_o = c + v_r

v_r = c(f/f_o - 1)

So, plugging in the relevant values, we have;

v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]

v_r = 3 × 10⁸(0.189)

v_r = 5.67 x 10⁷ m/s

Converting to mi/hr, 1 m/s = 2.23694 mile/hr

So, v_r = 5.67 × 10⁷ × 2.23694

v_r = 1.268 × 10⁸ mi/hr

5 0
4 years ago
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