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1 year ago

Why do electrons flow threw a wire

Physics

1 answer:

Zolol [24]1 year ago

3 0

Answer:

When electric voltage is applied, an electric field within the metal triggers the movement of the electrons, making them shift from one end to another end of the conductor. Electrons will move toward the positive side.

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The magnitude of the voltage induced in a conductor moving through a stationary magnetic field depends on the _______ and the __

Virty [35]

The correct answer is: Option (D) length, speed

Explanation:

According to Faraday's Law of Induction:

ξ = Blv

Where,

ξ = Emf Induced

B = Magnetic Induction

l = Length of the conductor

v = Speed of the conductor.

As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed

3 0

3 years ago

Read 2 more answers

What is a element that tends to be shiny, easy shaped, and a good conductor of heat?

ozzi

The answer to your question is Metal

8 0

3 years ago

Ghella [55]

Given parameters:

Mass of the body = 200g

Force on the body = 10N

Unknown parameters:

Acceleration produced by the force = ?

To solve this problem we must first define force in terms of mass and acceleration. This is possible due to the Newton's first law of motion.

Force = mass x acceleration

Here the unknown is acceleration and we can easily solve for it.

But we must take the mass to kilogram in order for it to cancel out.

1000g = 1 kg

200g = x kg = $\frac{200}{1000}$ = 0.2kg

Now input the parameters and solve;

10 = 0.2 x acceleration

Acceleration = $\frac{10}{0.2}$ = 50m/s²

The acceleration produced by the body is 50m/s²

4 0

3 years ago

A catapult with a spring constant of 10,000 N/m is used to launch a target from the deck of a ship. The spring is compressed a d

Mnenie [13.5K]

Answer:

Explanation:

Given;

spring constant of the catapult, k = 10,000 N/m

compression of the spring, x = 0.5 m

mass of the launched object, m = 1.56 kg

Apply the principle of conservation of energy;

Elastic potential energy of the catapult = kinetic energy of the target launched.

¹/₂kx² = ¹/₂mv²

where;

v is the target's velocity as it leaves the catapult

kx² = mv²

v² = kx² / m

v² = (10000 x 0.5²) / (1.56)

v² = 1602.56

v = √1602.56

v = 40.03 m/s

v ≅ 40 m/s

Therefore, the target's velocity as it leaves the spring is 40 m/s

6 0

2 years ago

You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with

Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

Explanation:

For this exercise we use the constructor equation or Gaussian equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

m = $\frac{h'}{h}$ = - $\frac{q}{p}$

h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

$\frac{1}{q} = \frac{1}{f } - \frac{1}{p}$

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

h ’= h q / p

where p has a high value and is the same for all systems

h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

f = 12 cm h ’= fo 12 cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0

3 years ago