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3 years ago

What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?

Physics

1 answer:

Schach [20]3 years ago

6 0

Answer:

Explanation:

Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.

This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.

2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.

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A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:

Ilia_Sergeevich [38]

Answer:

6.198 m/s²
4.48 s
453.77 m

Explanation:

5 0

3 years ago

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the

aleksandrvk [35]

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

$\theta= \frac{\omega_f-\omega_i}{2}\times t$

putting values

$\theta= \frac{0-2}{2}\times5$ = 5 rad

8 0

3 years ago

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s

Explanation:

It is given that,

Mass of the car, $m_1=1.1\times 10^3\ kg$

Mass of the truck, $m_2=2.3\times 10^3\ kg$

Initial velocity of the car, $u_1=22\ m/s$

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2$

$v_2=15.7\ m/s$

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0

3 years ago

A runner begins from rest at the starting line and travles for 6.5 seconds, a runner reaches a speed of 13.4 m/s what is the run

Butoxors [25]

The acceleration of the runner in the given time is 2.06m/s².

Given the data in the question;

Since the runner begins from rest,

Initial velocity; $u = 0$

Final velocity; $v = 13.4m/s$

Time elapsed; $t = 6.5s$

Acceleration of the runner; $a = \ ?$

<h3>Velocity and Acceleration</h3>

Velocity is the speed at which an object moves in a particular direction.

Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion

$v = u + at$

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

To determine the acceleration of the runner, we substitute our given values into the equation above.

$v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2$

Therefore, the acceleration of the runner in the given time is 2.06m/s².

Learn more about Equations of Motion: brainly.com/question/18486505

3 0

2 years ago

Ml(d^2θ/dt^2) =-mgθ

Nata [24]

The equation of motion of a pendulum is:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,$

where $\ell$ it its length and $g$ is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles ( $\theta \ll 1$ ), we can use:

$\sin\theta \simeq \theta.$

Additionally, let us define:

$\omega^2\equiv\dfrac{g}{\ell}.$

We can now write:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.$

The solution to this differential equation is:

$\theta(t) = A\sin(\omega t + \phi),$

where $A$ and $\phi$ are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.$

This justifies that the period depends only on the pendulum's length.

4 0

3 years ago