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3 years ago

How does charging by conduction occur?

Physics

1 answer:

lutik1710 [3]3 years ago

8 0

Answer : (A) A charged object is brought near a neutral object without touching it.

Explanation:

Bodies can be charged by the method of conduction. By conduction the body acquires the same charge as on the charging body.

Charging the body can be understood by the following example of charging a paper cylinder by conduction. Make a paper cylinder by rolling a strip of paper on a pencil and then gently pulling out the pencil. Suspend the paper cylinder by a string tied to its center, now touch the paper cylinder with a glass rod rubbed with silk so it has a positive charge. Remove the class rod and the again bring it near to the paper cylinder.

The paper cylinder is repelled by the positively charged rod. And when you bring a negatively rubbed rod, next to the cylinder it will be attracted towards the rod. This means that the paper cylinder has acquired a positive charge as on the glass rod due to conduction.

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The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to $1.129m^2$

(b) Charge on the capacitor is equal to $1.5\times 10^{-6}C$

Explanation:

We have given spacing between the plates d = 0.05 mm = $5\times 10^{-5}m$

Value of capacitance $C=1\mu F=10^{-6}F$

(A) Capacitance of a parallel plate capacitor is equal to $C=\frac{\epsilon _0A}{d}$

So $10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}$

$A=1.129m^2$

So surface area of the plate will be equal to $1.129m^2$

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to $Q=CV$

So $Q=1.5\times 10^{-6}C$

5 0

2 years ago

When something is hit harder how does the transverse wave change?

Flura [38]

When something is hit harder just like when sound is turned up the waves become higher and more frequent like a zig zag more so then wavy.

5 0

2 years ago

A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3

Natali5045456 [20]

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

ρ = 7.367 g / cm³

Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>

4 0

2 years ago

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago

How much ice (at 0Â°C) must be added to 1.90 kg of water at 79 Â°C so as to end up with all liquid at 8 Â°C? (ci = 2000 J/(kg.Â°

muminat

m = mass of the ice added = ?

M = mass of water = 1.90 kg

$c_{w}$ = specific heat of the water = 4186 J/(kg ⁰C)

$c_{i}$ = specific heat of the ice = 2000 J/(kg ⁰C)

$L_{f}$ = latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

$T_{ii}$ = initial temperature of ice = 0 ⁰C

$T_{wi}$ = initial temperature of water = 79 ⁰C

T = final equilibrium temperature = 8 ⁰C

using conservation of heat

Heat gained by ice = Heat lost by water

m $c_{w}$ (T - $T_{ii}$ ) + m $L_{f}$ = M $c_{w}$ ( $T_{wi}$ - T)

inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg

4 0

3 years ago