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a_sh-v [17]
3 years ago
15

Give two alterations in a generator to produce more electromotive force​

Physics
2 answers:
elixir [45]3 years ago
6 0

Answer:

Two ways by which the emf in an AC generator can be increased:

(i)The speed of the rotation of the coil is increased.

(ii)More powerful magnet is used to increase the strength of the magnetic field.

그것이 당신에게 도움이되기를 바랍니다 :)

가장 똑똑한 것으로 표시하십시오 :)

좋은 하루 되세요 :)

Mrac [35]3 years ago
4 0

Answer:

Rotating the coil faster

Increasing the number of coils

Using stronger magnets

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A block has two strings attached to it on opposite ends. One string has a force of 5 N,
juin [17]

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = <em>m</em> (1 m/s²)

where <em>m</em> is the mass of the block. Solve for <em>m</em> :

7 N = <em>m</em> (1 m/s²)

<em>m</em> = (7 N) / (1 m/s²)

<em>m</em> = 7 kg

(B) The friction force is proportional to the normal force, so that if <em>f</em> is the mag. of friction and <em>n</em> is the mag. of the normal force, then <em>f</em> = <em>µ</em> <em>n</em> where <em>µ</em> is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. <em>w</em>) cancel out:

<em>n</em> + (-<em>w</em>) = 0

<em>n</em> = <em>w</em>

<em>n</em> = <em>m</em> <em>g</em>

where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity.

<em>n</em> = (7 kg) (9.8 m/s²)

<em>n</em> = 68.6 N

Then

3 N = <em>µ</em> (68.6 N)

<em>µ</em> = (3 N) / (68.6 N)

<em>µ</em> ≈ 0.044

4 0
3 years ago
Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,
Leviafan [203]

Answer:

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

Explanation:

As we know that the charge per unit length of the long cylinder is given as

\lambda = \frac{q}{L}

here we know that the electric field between two cylinders is given by

E = \frac{2k\lambda}{r}

now we know that electric potential and electric field is related to each other as

\Delta V = - \int E.dr

\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr

\Delta V = -2k \lambda ln(\frac{b}{a})

\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

7 0
3 years ago
Compressional forces within the crust can produce:
brilliants [131]
<span>tension, compression, and shearing and can i get brainliest plz</span>
4 0
3 years ago
What is the approximate value of the gravitational force between a 73 kg astronaut and a 7.1×104 kg spacecraft when they're 89 m
Luden [163]

Answer:

F = 4.3671 * 10^{-8}\ Newtons

Explanation:

The gravitational force between two corpses is given by the following equation:

F = GMm/d^2

Where F is the force, G is the gravitational constant

(G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}), M and m are the masses of the corpses and d is the distance between them.

So we have that:

F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2

F = 4.3671 * 10^{-8}\ Newtons

5 0
3 years ago
a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem
VikaD [51]
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j 
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

6 0
3 years ago
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