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Dvinal [7]
3 years ago
12

The core of a 400 Hz aircraft transformer has a net cross-sectional area of 13 cm2. The maximum flux density is 0.9 T, and there

are 70 turns in the secondary coil. What is most nearly the rms voltage induced in the secondary coil
Physics
1 answer:
jenyasd209 [6]3 years ago
4 0

Answer:

32.76 Volt

Explanation:

frequency, f = 400 Hz

Area of crossection, A = 13 cm²

Maximum flux density, B = 0.9 tesla

Number of turns in secondary coil, N = 70

Let the maximum induced voltage is e.

According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.

e = dФ/dt

e=\frac{NBA}{t}

Time is defined as the reciprocal of frequency.

So, e = N B A f

e = 70 x 0.9 x 13 x 10^-4 x 400

e = 32.76 volt

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Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.

Explanation: Please see the attachments below

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3 years ago
When a current is passed through plasma and that current is increased will the temperature of the plasma increase . if it does t
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The temperature of the plasma will increase as current is passed through it.

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Current can be defined as a flow of charged particles through a medium and the particles includes: electrons or ions.

Temperature is the degree of hotness or coldness experienced by a material.

The more current passes through the plasma the more energy is induced making the particles to move faster which in turns causes temperature to increase and reducing the resistance of the plasma.

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6 0
1 year ago
Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far
olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

5 0
3 years ago
A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the buildin
Archy [21]

Answer:

19.3m/s

Explanation:

Use third equation of motion

v^2-u^2=2gh

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s

4 0
2 years ago
A baseball pitcher throws a 0.14 kg ball toward a batter who is 18 m away.
Olin [163]

Answer:

A. 112 J

Explanation:

KE = ½mv² = ½(0.14)40² = 112 J

5 0
3 years ago
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