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aleksklad [387]
3 years ago
12

Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica

l force of repulsion between these charges?
Physics
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant:  k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

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Answer:

the angle of reflection equals the angle of incidence—θr = θi. The angles are measured relative to the perpendicular to the surface at the point where the ray strikes the surface.

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 11.0 s. At th
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Answer

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

  ωf - ωi = α t

  11 α = 6 - 0

      = 0.545 rev/s²

The angular displacement

  θ₁= ωi t + (1/2) α t²

  θ₁= 0 + (1/2) (0.545)(11)^2

  θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

  ωf - ωi = α t

  14 α = 0 - 6

        = - 0.428 rev/s²

The angular displacement

  θ₂= ωi t + (1/2) α t²

  θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

  θ₂= 42 rev

total revolution in 25 s is equal to

θ =  θ₁ +  θ₂

θ =  33 + 42

θ = 75 rev

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3 years ago
How much time will it take a car traveling at 88 km/hr (55 mi/hr) to travel 500 km?
Alexeev081 [22]
The answer is5.68 hours
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3 years ago
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
serious [3.7K]

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

                                                                 = ± 11.0625 x 10^-6 C/m^2  

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C

7 0
3 years ago
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