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Paha777 [63]
1 year ago
12

Explain numericals of electricity chapter of class 10 and also diagrams.What is ammeter?What is voltmeter?potential difference?O

hm's law?Resistance?Electric circuitElectric current
Physics
1 answer:
SVEN [57.7K]1 year ago
3 0

Let's explain the following electrical terms.

• Electric current, can be said to be the flow of electric charge. These charges are through a conductor by moving electrons.

• Ammeter, is an instrument used to measure the amount of electric current in a circuit.

The name ammeter was derived from the unit of electric current (Amperes).

• Voltmeter, can be said to be an instrument used to measure the voltage(potential difference) in a circuit. It measures the electric potential between two points in an electrical circuit. It can also be called voltage meter.

• Electric circuit, can be defined as the conductive path for the flow of electric current.

It allows electric charge carriers to flow continuously.

• Resistance ,can be said to be the property of an electrical conductor which resists the flow of electric current. It is measured in ohms.

• Ohms law, states that the potential difference (V) between two points is directly proportional to the electric current across two points.

It is deonted as: V = I x R

Voltage = Current x Resistance

• Electric potential ,can be said to be the amount of electric potential energy at a point.

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Importance of liquid pressure
trapecia [35]

Answer:

hope it helps...

Explanation:

Both the water in the ocean and the air in the atmosphere exert pressure because of their moving particles. ... This causes greater pressure. Denser fluids such as water exert more pressure than less dense fluids such as air. The particles of denser fluids are closer together, so there are more collisions in a given area.

7 0
3 years ago
Other quanto
Alex73 [517]

I'm not sure what you were trying to put here

5 0
3 years ago
If there is a huge boulder on your lawn and you want to determine its density, but it is
Olin [163]

There are different options here but all of them work by approximating and assuming.

i) that the boulder is above ground.

ii) that the bottom surface of the boulder is known.

iii) the shape of the boulder is taken into account.

The most accurate way is measuring it by displacement method but the boulder is immovable hence the volume can be calculated by measuring the boulder or a waterproof box to be built around the boulder and calculate the volume occupied by boulder.

All the above methods are estimating methods.

*Another way to find the density is through specific gravity.

S.G = <u>Density</u><u> </u><u>of</u><u> </u><u>object</u>

Density of water

If the material that makes the boulder is known that is if it's stone or a mineral then the specific gravity can be found.

If the boulder is purely rock then S.G lies between 3 - 3.5 and the density of water is known thus the density of the boulder can be found without moving the boulder.

This is what I think after correction and allthe best!

3 0
2 years ago
Read 2 more answers
Using the right amount of significant figures, calculate the answer to the following problem, 15.33879 + 15.555
MAXImum [283]

Answer:

there are 7 significant figures

Explanation:

15.33879+15.555

=30.89379

there are 7 significant figures

mark me as brainliest plyyzzz

7 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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