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AlekseyPX
1 year ago
8

Solenoid is wound with 2000 turns per meter when the current is 5.2 a, what is the magnetic field within the solenoid?

Physics
1 answer:
Drupady [299]1 year ago
7 0

The magnetic field within the solenoid is 0.013T

<h3>What is Solenoid ?</h3>

An electromagnet known as a solenoid creates a regulated magnetic field using a helical coil of wire whose length is significantly higher than its diameter. When an electric current is sent through the coil, it may create a consistent magnetic field inside a defined region of space.

A solenoid operates by creating an electromagnetic field surrounding an armature, which is a moving core. The electromagnetic field causes the armature to move, and when it does, it opens and closes valves or switches, converting electrical energy into mechanical motion and force.

The magnetic field within the solenoid can be calculated by the expression given below.

Write the expression for magnetic field within the solenoid.

B = u₀nI

B = 4π * 10⁻⁷ * 2000 * 5.2

= 0.013 T

to learn more about solenoid go to -

brainly.com/question/1873362

#SPJ4

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The power output of a laser is measured by its wattage, the number of joules of energy it radiates per second (1 W = 1 J s-1). A
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Explanation:

It is given that,

Wavelength of red light, \lambda=676.4\ nm=676.4\times 10^{-9}\ m

Power of the laser, P=300\ mW=0.3\ W

(a) The energy carried by each photon is given by :

E=\dfrac{hc}{\lambda}

E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{676.4\times 10^{-9}}  

E=2.94\times 10^{-19}\ J

(b) Let n is the number of photons emitted by the laser per second. It can be calculated as :

n=\dfrac{Power}{Energy\ of\ one\ photon}

n=\dfrac{0.3}{2.94\times 10^{-19}}      

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Hence, this is the required solution.    

7 0
3 years ago
An egg is dropped from a building that is 61 m high.
Allisa [31]

Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

time = 3.5 s

Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

<u>Initial Velocity = 0 m/s</u>

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

<u>t = 3.5 s</u>

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.6 m/s

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3 years ago
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A track star in the broad jump goes into the jump at 12 m/s and launches himself at 20° above the horizontal. How long is he in
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Explanation:

It is given that,

Initial speed of the broad jump, u = 12 m/s

It is launched at an angle of 20 degrees above the horizontal. Let t is the time for which the track star i in the air before returning to Earth. The motion of the track star in the broad jump can be treat as the projectile motion. The time of flight of the projectile is given by :

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Putting all the values in above equation as :

t=\dfrac{2\times 12\times \ sin(20)}{9.8}

t = 0.837 seconds

So, the time for which the track star is in air is 0.837 seconds. Hence, this is the required solution.

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