All categories

3 years ago

In order to induce electrical energy into a conductor or generator, what three factors must be present?

1 answer:

6 0

The electricity produced by a generator works through the concepts of the following three factors; magnetic field, voltage and current. When a conductor with current flowing in it is placed in a magnetic field, it will cause the electrons to move in a direction perpendicular to the magnetic field. When working with generators, the guide for this direction is Fleming's Right Hand Rule. Since the electron move in perpendicular motion with magnetic field all the time, it would cause it to spin in a helical direction. These turns would then induce voltage and create electricity.

You might be interested in

if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp

Vilka [71]

Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

$\displaystyle{\sin \theta = \dfrac{u_y}{u}}$

$\displaystyle{u_y}$ is vertical component of initial velocity and $\displaystyle{u}$ is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so $\displaystyle{\theta}$ = 50°. Substitute in the formula:

$\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}$

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0

2 years ago

Four children pull on the same stuffed toy at the same time yet there is no net force on the toy. How is this possible.

Stels [109]

Answer:

Net force is Zero.

Explanation:

If all forces that are equal and opposite are exerted on an object the resulting force will be Zero.

3 0

3 years ago

Read 2 more answers

When a object slows down what direction is the force going?

trapecia [35]

Answer: when a object gets slowed down, it's force is either going into friction and drag, if it's on the ground, and weight+drag+friction, if it's in the air.

Explanation:

7 0

3 years ago

A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p

Anna71 [15]

Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

final velocity v = 6 m /s

acceleration due to gravity g = 9.8 m /s²

v² = u² + 2 g h

6² = 0 + 2 x 9.8 x h

h = 1.837 m .

For second case u = 3 m /s

v² = u² + 2 gh

= 3² + 2 x 1.837 x 9.8

= 9 + 36

= 45 m

v = 6.7 m /s

8 0

3 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago