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daser333 [38]
2 years ago
5

a generator uses loops of area 0.239 m^2, rotating 373 rad/s in a 0.0639 t magnetic field. how many loops must the coil have to

generate a maximum emf of 2500 v?
Physics
1 answer:
Kaylis [27]2 years ago
6 0

The loops must the coil have to generate a maximum emf of 2500 will be 439.

<h3 /><h3>What is the faraday law of electromagnetic induction?</h3>

According to  Faraday's law of electromagnetic induction, the rate of change of magnetic flux linked with the coil is responsible for generating emf in the coil resulting in the flow of amount of current.

Given data;

Area,A = 0.239 m²

Angular velocity,ω=373 rad/sec

Magnetic field,B=0.0639 T

Maximum emf,E= 2500V

The formula for the maximum induced voltage is;

E{max} = N ×  B × A × ω

2500 = N × 0.639 × 0.0239 × 373

N = 438.66

N = 439 \ turns

Hence, loops must the coil have to generate a maximum emf of 2500  will be 439.

To learn more about the faraday law of electromagnetic induction refer to;

brainly.com/question/26334813

#SPJ1

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An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
UkoKoshka [18]

An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

  • The single valance electron leads to weak bonds amongst the element's atoms which makes them soft
  • The elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.

Examples of alkali electrons include:

  • Lithium
  • Sodium
  • Potassium etc

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

<em>Find out more at brainly.com/question/18722874.</em>

6 0
3 years ago
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A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round
Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s

v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s

Now we find the centripetal acceleration which is given by

a_c=\frac{v^2}{r}

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2

we also have a tangential acceleration, which is given by

a_t = \frac{v-u}{t}

where

t = 17.0 s

Substituting values,

a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2

6 0
3 years ago
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A CD has to rotate under the readout-laser with a constant linear velocity of 1.25 m/s. If the laser is at a position 3.7 cm fro
Savatey [412]

Answer:N=322.53 rpm

Explanation:

Given

Linear velocity (v)=1.25 m/s

Position from center is 3.7 cm

we know

v=\omega \times r

1.25\times 100=\omega \times 3.7

\omega =\frac{125}{3.7}=33.78

and \frac{2\pi N}{60}=\omega

N=\frac{\omega \times 60}{2\pi }

N=\frac{33.78\times 60}{2\pi }

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8 0
3 years ago
T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?
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Answer:

P_J=195N

Explanation:

From the question we are told that

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Generally the equation for momentum is mathematically given by

P=mv

Therefore

T-Joe momentum P_J

P_J=65*3

P_J=195N

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Which properties do metalloids share with metals?
Inga [223]
Metals are not brittle so it can’t be the first one or the third one, both metalloids and metals are shiny so it can’t be the second one. Therefore, it would be the last one because both metalloids and metals are shiny and both are solids at room temperature because it is not a high enough melting point.

ANSWER: Both are shiny and are solid at room temperature.
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