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4 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.

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it takes a school bus full of kids longer to stop than it does a small car with a single passenger. which law is this an example

exis [7]

Newton´s second law of motion

7 0

4 years ago

Two masses are separated by a distance r. If this distance is doubled, is the force of interaction between the two masses double

baherus [9]

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

This is mathematically represented as

where F is the force acting between the charged particles

r is the distance between the two charges measured in m

G is the gravitational constant which has a value of <em>6.674×10^-11 Nm^2 kg^-2</em>

m1 and m2 are the masses of the objects measured in Kg

Now if the distance between the is doubled then r becomes 2r. Substituting this in the above formula we get the new Force as

Force (new) = (G X m1 x m2) /(2r)∧2

Thus dividing Force(new)/Force we get

Force(new)/Force = 1/4.

Thus the gravitational force becomes 1/4th of the original value if the distance between the two masses are doubled.

7 0

3 years ago

Read 2 more answers

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 375 N

spayn [35]

Answer:

The conservation of energy should be used to answer this question.

At the position where the spring is unstretched, the elastic potential energy of the spring is zero.

$K_1 + U_1 - W_f = K_2 +U_2\\K_1 - W_f = U_2$

since $U_1$ and $K_2$ is equal to zero.

$W_f = F_fx\\\\U_2 = \frac{1}{2}kx^2\\\\19 - (10)x = \frac{1}{2}(375)x^2\\\\375x^2 + 20x - 38 = 0$

The roots of this quadratic equation can be solved by using discriminant.

$\Delta = b^2 - 4ac\\x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$

$x_1 = -0.346\\x_2 = 0.292$

We should use the positive root, so

x = 0.292 m.

We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.

$K_2 + U_2 - W_f = K_3 + U_3\\U_2 - W_f = K_3$

since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.

$\frac{1}{2}kx^2 - F_fx = K_3\\K_3 = 15.987 - 2.92 = 13.067 J$

Explanation:

In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.

In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).

8 0

3 years ago

I would like to know the correct answer for this

Mnenie [13.5K]

The last choice on the list is the correct one, for both #2 and #3.

3 0

4 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago