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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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The type of material of which the resistor is made.

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The thickness of the resistor.

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What type of energy is involved when a river moves sediment and erodes its banks?
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3 years ago
Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom
Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i = \dfrac{1}{2}mv_f^2

             v_f = \sqrt{2gh_i}

                   = \sqrt{2gL(1- cos\theta)}

                   = \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}

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the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2

          gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2

             v_f = \sqrt{v_1^2+2gh_i}

             v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}

                       v_f= 11.29 m/s

3 0
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I apologize for the messy graph in advance. If you have any further questions please let me know

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Kazeer [188]

Answer:

False

because I got that question and I gotted right

6 0
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