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3 years ago

Which state of matter would be described as a highly energized charge particles with moving extremely fast

Physics

1 answer:

Leni [432]3 years ago

8 0

Answer:

plasma

Explanation:

gegehhehehhrhrh

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The distance between two parallel wires carrying currents of 10 A and 20 A is 10 cm. Determine the magnitude and direction of th

Marianna [84]

Explanation:

It is given that,

Current in wire 1, I₁ = 10 A

Current in wire 2, I₂ = 20 A

Distance between wires, d = 10 cm = 0.1 m

Force per unit length is given by :

$\dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

$\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 10\times 20}{2\pi \times 0.1}$

$\dfrac{F}{l}=0.0004\ N/m$

$\dfrac{F}{l}=4\times 10^{-4}\ N/m$

So, the magnetic force acting per unit length of the wires $4\times 10^{-4}\ N/m$ . Since, the current is in same direction. So, the force is attractive in nature.

6 0

3 years ago

The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force

yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given

N = W Cos30

N = (500) Cos30

N = 433 N

3 0

3 years ago

Light is:<br><br> a wave<br> a particle<br> neither

valentina_108 [34]

A wave !!!! it’s a type of energy that is called electricmagnetic energy

8 0

3 years ago

Read 2 more answers

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

4 0

3 years ago

Technician A says that magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them ove

Eva8 [605]

Answer:

Both A and B are correct

Explanation:

Magnetic line of force can be seen by placing iron fillings on a piece of paper and then hold a magnet over them, the iron fillings align themselves in the magnetic field of the magnetic along the line of forces thus showing the loops of the magnetic line of forces.

Also, we know that unlike poles attract where as like poles repel each other. This fact can be proved when we hang a bar magnet it aligns itself in the North-South direction because of the the magnetic field due to the alignment of the geographical North-South poles and the poles of the bar magnet and the compass works the same way.

Thus both the technicians are correct.

4 0

3 years ago

Read 2 more answers