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yawa3891 [41]
4 years ago
8

A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha

t has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet?
Physics
1 answer:
PilotLPTM [1.2K]4 years ago
8 0

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

F_{e} = mg

where

F_{e} = the force between the electric field

m = mass

g = gravitational force

F_{e} =qE

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

F_{e} =qE

mg = qσ/2ε₀

σ = (2mg ε₀)/q

  = 2 (0.016) (9.8)  (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

  = - 3.23 x 10⁻⁷ C/m²

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Thermal energy transformation is taking place.
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Find the weight of a 4.2 kg backpack,<br> A) 41.16 N<br> B) 0.42 kg<br> C) 41.16 kg<br> D) 0.42 N
Olin [163]

Answer:

Although the weight is given and we are asked for the weight, the odd wording of the question implies that we are being asked for the gravitational force on the backpack.

If it is gravitational force that is being requested, then the correct answer is A) 41.16 Newtons.

If it is actual weight being asked for, then none of the answers are correct, as the weight of 4.2 kg is given in the question, and does not match any of the answers.

6 0
3 years ago
The diagram below shows a subduction zone where oceanic crust is sinking into the mantle underneath continental crust as two
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Metal ores

Explanation:

in an area where subduction has occurred in times past, metal ores are likely to be found.

Metallic ores find subduction zone regions very favorable to crystallize out of a magma.

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4 0
3 years ago
Jamie decides to drop an egg that has a mass of 345 g off the top of a 8.2 m building. How fast will it be falling right before
zlopas [31]

Answer:

13 m/s

Explanation:

Given:

Δy = 8.2 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (8.2 m)

v = 12.7 m/s

Rounded to two significant figures, the speed is 13 m/s.

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3 years ago
The electric potential is 40 V at point A near a uniformly charged sphere. At point B, 2.0 μm farther away from the sphere, the
KatRina [158]

Answer:

Explanation:

Potential at a point near a charge = Q / 4πε₀ R

where Q is charge given , R is distance of point from the charge .

Q / 4πε₀ R  = 40

Electric field E at A = -dV / dR

= .16 x 10⁻³ / 2 x 10⁻⁶

= .08 x 10³

= 80 N/C

E = Q / 4πε₀ R²

80 =  Q / 4πε₀ R x R

80 = 40 / R

R = 40 / 80

= .5 m .

4 0
3 years ago
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