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3 years ago

What is the difference between current electricity and static electricity

Physics

2 answers:

Allisa [31]3 years ago

7 0

Answer:

The most significant difference between the static and current electricity is that in static electricity the charges are at rest and they are accumulating on the surface of the insulator. Whereas in current electricity the electrons are moving inside the conductor.

Mama L [17]3 years ago

3 0

Answer:

Static electricity is electricity that does not flow. It is produced by electrons rubbing off one object and being collected on another. It cannot flow through a wire like the electricity in your home. An example of static electricity is rubbing your hair on a balloon. static electricity causes your hair to stick up. On the other hand, current electricity can flow. It is generated by batteries and power plants. In current electricity the electrons are moving inside the conductor. Current electricity runs through circuits. These circuits run all around your homes. Current electricity is also what we use in our homes. An example of current electricity is the electricity that we use from outlets.

Explanation:

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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers

How does the elbow medical and lateral epicondylitis?

dlinn [17]

Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. The bony bump on the outside (lateral side) of the elbow is called the lateral epicondyle. The ECRB muscle and tendon is usually involved in tennis elbow. 
Medial epicondylitis, or “golfer’s elbow,” is an inflammation of the tendons that attach your forearm muscles to the inside of the bone at your elbow. It's identified by pain from the elbow to the wrist on the inside (medial side) of the elbow. The pain is caused by damage to the tendons that bend the wrist toward the palm.

4 0

3 years ago

When does a scientist create the conclusion?

Lerok [7]

After they have gathered enough information by testing the theory.

7 0

3 years ago

A need the answers pleaseeeee

IRISSAK [1]

It's just asking you to sit down and COUNT the little squares in each sector.

It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it. That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.

(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)

7 0

3 years ago

a piece of metal with a mass of 15.3 grams has a temperature of 50.0°C. When the metal is placed in 80.2 grams of water at 21.0°

AleksAgata [21]

Answer:

1.21

Explanation:

Heat rise in the body happens due to heat supplied by water to the body.

Heat rise in body = m₁ c₁ ΔT₁

Where m₁ is mass of body and c₁ is its specific heat of body

Heat lost from water to the body = m₂ c₂ ΔT₂

Where m₂ is mass of water and c₂ is its specific heat of water ( c₂ =1 (since water))

Equating both:

15.3 x c₁ x 4.3 = 80.2 x 1 x 4.3

⇒ c₁ = 80.2 / (15.3 x 4.3) = 1.21

6 0

3 years ago

Read 2 more answers