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3 years ago

How does charging by conduction occur?

Physics

2 answers:

Aloiza [94]3 years ago

7 0

Charging by conduction involves the contact of a charged object to a neutral object.

C) A charged object touches a neutral object

tia_tia [17]3 years ago

4 0

Answer:

Option (C)

Explanation:

When a charged body is brought in contact to uncharged body, the process of conduction takes place.

For example, there are two identical bodies one having charge 10 C and another is uncharged, when they kept in contact to each other the charges will be distributed.

It means after conduction the charge on both the bodies will be 5 C.

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ziro4ka [17]

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7 0

3 years ago

An electrical circuit is powered by a 1.5-volt battery and contains a light bulb producing 5 Ohms of resistance. There is also a

Fed [463]

Answer:

if the resistor is fitted in series with the bulb , then the current flowing will be 0.15 A.....

if the resistor if fitted in parallel with the bulb ,

the current flowing will be 0.6 A

Explanation:

total potential difference = 1.5V

when resistors in series ,

total equivalent resistance is = 5 + 5 = 10 ohm

so current = 1.5 ÷ 10 = 0.15

when resistors in parallel ,

total equivalent resistance is = (5 × 5)÷(5 + 5) =2.5 ohm

so current = 1.5÷2.5 = 0.6 A

8 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$