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3 years ago

An electrical current flowing through a filament bulb causes it to get hot. Explain why this

Physics

1 answer:

sergij07 [2.7K]3 years ago

7 0

Answer:

The reason the filament heats up is because it has a high resistance, which means that as electrons move through the filament, they lose a lot of energy.

First, what is current? Current is comprised of electrons moving through an electric field from a high electric potential to a lower potential. For the current to decrease then, something would need to happen to the electrons that go into the light bulb. If 1 electron goes into the light bulb, then at the end of everything I need to still have 1 electron someplace. So how do electrons passing through the bulb make light?

Incandescent light bulbs have a small filament which when heated begins to glow and emit light. The reason the filament heats up is because it has a high resistance, which means that as electrons move through the filament, they lose a lot of energy. You can think of it as walking on a sidewalk compared to walking in waist deep water. A wire is like a sidewalk. It has some resistance, but it is so tiny that it can generally be ignored which is why wires are useful in electronic circuits. The high resistance of the light bulb is like trying to walk through waist deep water. Here energy is being taken from the electrons because of the interactions with the atoms in filament which causes those atoms to heat up, which in turn makes them emit light.

The light bulb is not doing anything to the electrons, so we expect then that any electrons going into the bulb should come out the other side. Since current is just flowing electrons, current stays the same.

Since current is the same on both sides, we know that the electrons are all moving together. Think of it like being in a big loop of people. Since everyone is in a big line you could imagine that you could only move as fast as the slowest person in the line. If everyone is on a big loop of sidewalk then everyone could run around in a circle. This is like having a large current in a loop of wire, or what we call a short. To put the equivalent of a lightbulb into our human circuit, imagine that one section of the sidewalk dips into a pool of water. Now everyone is stuck going as fast as the people trudging through the water. This is why current everywhere in a circuit is smaller when a resistor is introduced. As people trudge through the water they have to work hard to get through the water and they use energy. In a circuit, this energy comes from the voltage source, like a battery. The battery loses energy because it has to "pull" the electrons through the high resistance, and this is why the voltage drops across the light bulb

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I WILL GIVE BRAINLIST just please help with these problems 3

statuscvo [17]

Answer:

16. 68.18 Km/h

17. 3 miles.

Explanation:

16. Determination of the speed

Distance travelled = 150 Km

Time = 2.2 hours

Speed =?

Speed is simply defined as the distance travelled with time. Mathematically, it is expressed as:

Speed = Distance / time

With the above formula, we can obtain the speed as follow:

Distance travelled = 150 Km

Time = 2.2 hours

Speed =?

Speed = Distance /time

Speed = 150 / 2.2

Speed = 68.18 Km/h

17. Determination of the distance.

Speed = 3 mph

Time = 1 hour

Distance =?

Speed = Distance /time

3 = distance / 1

Distance = 3 miles

3 0

3 years ago

sweet-ann [11.9K]

Answer:

Explanation:

1.) The reaction is at dynamic equilibrium.

A: Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

2.) Which statement about the reaction is necessarily correct?

A: Both calcium carbonate and sodium carbonate are being produced.

3.) Both calcium carbonate and sodium carbonate are being produced.

A: The reaction is reversible.

4.) What is the fastest motion that can be measured in any frame of reference?

A: 300,000 km/s

5.) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north.

Why are we able to use these motions to describe the motion relative to the ground?

A: The people are moving much slower than the speed of light so the ground acts as a frame of reference.

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3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t

elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

Angular velocity ( $\omega$ ) = 2.23 rps

Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

= 2.23 revolutions per second

= $2.23 \times 2 \times 3.14 rad/s$

= 14.01 rad/s

Now, tangential speed will be calculated as follows.

Tangential speed, v = $R \times \omega$

= 0.379 x 14.01

= 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

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3 years ago

me (5) A taxi is travelling at 15. m/s. Its driver accelerates with acceleration 3 m/s for 4 s. What is its new velocity? A car

photoshop1234 [79]

Answer:

Solution

initial velocity = 15 m/s

acceleration = 3m/s^2

time = 4 s

final velocity = 15+3×4

=15+12

=27m/s

A car accelerates from 20m/s to 30m/s in 10 second

T9 be honest I think some part of the second question is missing

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3 years ago

A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the accele

ElenaW [278]

(a) $1.23 m/s^2$

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

$d = ut + \frac{1}{2}at^2$

Since u=0 (the block starts from rest), it becomes

$d=\frac{1}{2}at^2$

So by solving the equation for a, we find the acceleration:

$a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2$

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

$W_p = mg sin \theta$

where

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

$\theta=33.0^{\circ}$ is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

$F_f = - \mu mg cos \theta$

where

$\mu$ is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

$W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma$

Solving for $\mu$ , we find

$\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50$

(c) 12.3 N

The frictional force acting on the block is given by

$F_f = \mu mg cos \theta$

where

$\mu = 0.50$ is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=33.0^{\circ}$ is the angle of the incline

Substituting, we find

$F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N$

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

$v^2 - u^2 = 2ad$

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

$v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s$

5 0

3 years ago