How is static electricity different from electric current

Physics

1 answer:

Sindrei [870]3 years ago

8 0

Answer:

1. Static electricity is caused by the build up of electrical charges on the surface of objects, while current electricity is a phenomenon from the flow of electrons along a conductor.

Explanation:

You might be interested in

fomenos

Answer:

The bulbs should be connected in parallel.

Explanation:

We want to find out a way to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other bulbs stays lit.

We must connect the two bulbs in parallel so that even when one bulb is burns out, it will have no effect on the other bulb and the 2nd bulb will keep on working. The current flowing in each bulb will depend upon the resistance of each bulb and the voltage will be same across each bulb.

On the other hand, if we use a series circuit then if one bulb burns out then the there is no flow of current in the circuit and therefore, the second bulb will not be operational.

The current flowing through each bulb is given by

I = V/R

The voltage across each bulb is given by

V = IReq

Where I is the current and Req is the equivalent resistance of the two bulbs connected in parallel and is given by

Req = (R₁*R₂)/(R₁+R₂)

The connection diagram is attached where two bulbs are connected in parallel and are power with a battery.

4 0

3 years ago

A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then

Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J