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sammy [17]
3 years ago
5

A 1 mg ball carrying a charge of 2 x 10-8 C hangs from a

Physics
1 answer:
Fed [463]3 years ago
7 0

Answer:

σ = 0.255*10^-3 C/m²

Explanation:

The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.

At equilibrium ,

Upward force = downward force

Tcosθ = mg ----(1)

Assuming that the Forward force = backward force, then

Tsinθ = σq/ε₀

[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)

Dividing equation (2) by (1)

Tsinθ/Tcosθ = qσ/ε₀mg

⇒Tanθ = qσ/ε₀mg

σ = ε₀mg tanθ/q

Now substituting the values of

σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8

σ = (8.85*10^-12 * 0.5774) / 2*10^-8

σ = 5.11*10^-12 / 2*10^-8

σ = 0.255*10^-3 C/m²

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If the fly experienced a force of 100 N when it hits how big was the force on the vehicle in this collision
iogann1982 [59]

Answer:

100N

Explanation:

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so if the fly experienced 100N, then the car will also experience 100N

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Which TWO correctly relate the attraction between the particles of a liquid and the temperature at which the liquid changes stat
Naddika [18.5K]
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3 years ago
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In a 35 mm single lens reflex camera (SLR) the distance from the lens to the film is varied in order to focus on objects at vary
Evgesh-ka [11]

Answer:

range of movement is 1.49 mm

Explanation:

given data

focal length = 45 mm

distance = 1.4 m

distance from the lens = 35 mm

distance from infinity down = 1.4 m =

to find out

range of movement

solution

we will apply here lens equation that is

1/f = 1/p + 1/q

here f = 45 and p =  infinity to 1400 mm

we find here image distance that is q

1/45 = 0 + 1/q             ......1

q = 45 mm

and

1/45 = 1/1400 + 1/q     ......2

q = 46.49

so range of movement

that is 46.49 - 45

range of movement is 1.49 mm

8 0
3 years ago
An object is placed 96.5 cm from a glass lens (n = 1.51) with one concave surface of radius 24 cm and one convex surface of radi
poizon [28]

Answer:

image is vertical at distance -203.62 cm

magnification is 2.110

Explanation:

given data

n = 1.51

distance u = 96.5 cm

concave radius r1 = 24 cm

convex radius r2 = 19.1 cm

to find out

final image distance and magnification

solution

we will apply here lens formula to find focal length f

1/f = n-1 ( 1/r1 - 1/r2)   .......................1

put here all value

1/f = 1.51 -1 ( -1/24 + 1/19.1)

f = 183.43

so from lens formula

1/f = 1/v + 1/u     .............................2

put here all value and find v

1/183.43 = 1/v + 1/96.5

so

v = −203.62 cm

so here image is vertical at distance -203.62 cm

and

magnification are = -v /u

magnification =  203.62 / 96.5

magnification is 2.110

3 0
4 years ago
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