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liraira [26]
2 years ago
12

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

ial difference would appear as:_________
Physics
1 answer:
vfiekz [6]2 years ago
5 0

The potential difference comes out to be

10 \times 10 {}^{ - 3} m

Given:

σ = 8. 85 × 10-9 c/m2

we know,

E = \frac{σ}{2ε0}

E =  \frac{8.85 \times 10 {}^{ - 9} }{2ε0}

E =  \frac{v}{d}

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

=  \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }

Δ = 10 \times 10 {}^{ - 3} m

Thus the potential difference is

10 \times 10 {}^{ - 3} m

Learn more about potential difference from here: brainly.com/question/28165869

#SPJ4

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4 0
3 years ago
The length of a certain wire is doubled and at the same time its radius is also doubled. What is the new resistance of this wire
Leto [7]

Answer:

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

Explanation:

Consider a wire with:

Resistance = R

Length = L

Area = A = πr²

where, r = radius

ρ = resistivity

Then:

R = ρL/A

R = ρL/πr²   --------------- equation 1

Now, the new wire has:

Resistance = R'

Resistivity = ρ

Length = L' = 2 L

Radius = r' = 2r

Area = πr'² = π(2r)² = 4πr²

Therefore,

R' = ρL'/πr'²

R' = ρ(2 L)/4πr²

R' = (1/2)(ρL/πr²)

using equation 1:

<u>R' = R/2</u>

<u>Therefore, the new resistance of the wire is twice the value of the initial resistance.</u>

4 0
3 years ago
A student on roller skates glides across a floor with a lunch tray. Total mass is 1.3 kg at waist level 1.1m above the floor fro
Anna71 [15]

Answer:

22.3 work is left to be done

2.4 work are done

8 0
2 years ago
An abstract sculpture consists of a ball (radius R = 76 cm) resting on top of a cube (each side L = 200 cm long). The ball and t
scZoUnD [109]

Answer:

132.9 cm

Explanation:

Data provided in the question:

Radius of the basll = 76 cm = 0.76 m

Side of the box = 200 cm = 2 m

Density of the ball and cube are equal

let the density be 'D'

Now,

Mass of ball, M = Volume × Density

= \frac{4}{3}\pi r^3  × D

= \frac{4}{3}\pi (0.76)^3× D

= 1.838D

Mass of cube, m = L³ × D

= 2³ × D

= 8D

Thus,

center of mass, y = [ My₁ + my₂ ] ÷ [M + m]

here,

y₁ = center of mass of ball with respect to floor

as the center mass of sphere lies at the center of the sphere

= Length of cube + radius of sphere

= 2 + 0.76

= 2.76 m

y₂ = Center of mass of cube = \frac{L}{2}=\frac{2}{2} = 1 m

Thus,

y = [ ( 1.838D × 2.76 ) + (8D × 1 ) ] ÷ [1.838D + 8D]

= 13.07288D ÷ 9.838D

= 1.329 m

or

= 132.9 cm

3 0
3 years ago
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