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malfutka [58]
3 years ago
12

Two point charges lie on the x axis. A charge of + 2.20 pC is at the origin, and a charge of − 4.80 pC is at x=−11.0cm. What thi

rd charge should be placed at x=+20cm so that the total electric field at x=+10.0cm is zero?
Physics
2 answers:
k0ka [10]3 years ago
4 0
As we know that electric field at any point caused due to a point charge given by, kQ/r^2

Where, k is dielectric constant of medium, Q is the point charge and r is the distance between the point charge and point where electric field is to be measured.

Let us assume a charge of q is placed at x=+20cm to make electric field at
x=+10cm zero.

Total electric field at x=+10cm can be shown by equation,

(k X 2.20 X 10^-12)/(10-0)^2  +  (k X -4.80 X 10^-12)/(10-(-11))^2  + (k X
q)/(10-20)^2  = 0

=> 2.20 X 10^-14 - 1.09 X 10^-14 + q X 10^-2 = 0

=> q = 1.09 X 10^-12 - 2.20 X 10^-12

=> q = -1.1 X 10^-12 Coulombs or -1.1 pC
My name is Ann [436]3 years ago
4 0

Answer: The charge must be -1.11pC

Explanation: The field done by a charge is equal to:

E(x) = k*q/(x - x')^2

Where k is the constant, q is the charge and x' is the position of the charge, then, the total field at the point x = 10cm must be also zero, we have:

E(10cm) = k*(2.20pC/(10cm - 0cm)^2 - 4.80pC/(10cm + 11cm)^2 + Q/(10cm - 20cm)^2) = 0

And we must find the value of Q.

(2.20pC/(10cm)^2 - 4.80pC/(21cm)^2 + Q/(-10cm)^2) = 0

Q = (-2.20pC/(10cm)^2 + 4.80pC/(21cm)^2)*(10cm)^2 = -1.11pC

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Alex

Answer:

Radio Wave

Micro Wave

Explanation:

Electromagnetic waves are transverse waves composed by the perpendicular oscillating electric and magnetic fields.

EM waves have both Electrical and magnetic features.

they travel in the velocity of light (3*10⁸ ms⁻¹)

Electromagnetic spectrum is obtained according to their wave length and the frequency. Due to wave length range it's categorized. Here is the decreasing  order of wave length and increasing order  of different wave types in electromagnetic spectrum

  • Radio Wave
  • Micro Wave
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5 0
3 years ago
Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with
ad-work [718]

Answer:

y₀ = 10.625 m

Explanation:

For this exercise we will use the kinematic relations, where the upward direction is positive.

         y = y₀ + v₀ t - ½ g t²

in the exercise they indicate the initial velocity v₀ = 8 m / s.

when the rock reaches the ground its height is zero

         0 = y₀ + v₀ t - ½ g t²

        y₀i = -v₀ t + ½ g t²

let's calculate

         y₀ = - 8  2.5 + ½  9.8  2.5²

         y₀ = 10.625 m

7 0
2 years ago
Please help me!! Need this done before the 40 min end
Alex_Xolod [135]

Answer:

its c

Explanation:

bc i know

5 0
3 years ago
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,
Ne4ueva [31]
A)  f = 1.8 rev/s = 2 Hz 
<span>T = 1 / f = 0.55s

B)  not really sure..srry

C)  </span><span>T = 2 pi √ ( L / g ) </span>
<span>0.57 = 2 x 3.14 x √ ( 0.2 / g )
</span><span>
g = 25.5 m/s²
</span>
Hope this helps a little at least.. :)

5 0
3 years ago
5. A man has a weight of 100 Newtons. How much work is done if he climbs 4 meters up a ladder? Plug numbers under the equation.
Virty [35]

Answer:

<h2>400 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 4

We have the final answer as

<h3>400 J</h3>

Hope this helps you

5 0
2 years ago
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