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3 years ago

What is the source of all electromagnetic waves?

Physics

1 answer:

kherson [118]3 years ago

6 0

Answer:

Accelerating charges.

Explanation:

Electromagnetic waves are waves produced by the vibration of both electrical and magnetic fields.

This interaction produces an energy source that does not require any medium to propagate.

To produce electromagnetic waves, electric and magnetic fields must be vibrating.

An electric charge produced when vibrating under voltage will produce electromagnetic waves. This is the same for all sources of these waves.

The sun produces electromagnetic waves. A lot of human activities also does this.

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What are the characteristics of a nebulae? (Select all that apply.)

erica [24]

Answer:

B. contain hydrogen

C. clouds of gas and dust

E. needed to create a star

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

Some of the examples of stars are; Vega, Sun (closest to planet Earth), Antares, Betelgeus, Canopus, etc.

Stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He). The chronological order in which the formation of a star occur are;

1. Gravity pulls gas and dust together to form dense cores.

2. A protostar forms as mass increases.

3. Nuclear fusion begins under high pressure.

Scientists have been able to understand and discover that, gravity pulled materials (low-density cloud of interstellar gas and dust known as a nebula) together forming the planetary bodies in our solar system.

A dark nebula can be defined as an interstellar cloud that is so dense as a result of high concentration of gas and dust and as such it obscures the visible wavelengths of light from stars behind it, thus appearing completely opaque (dark patch) in front of a bright emission nebula or in regions having plenty stars.

The characteristics of a nebulae are;

I. It contain hydrogen.

II. Clouds of gas and dust

III. It is needed to create a star.

7 0

3 years ago

A 51-g rubber ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact w

Simora [160]

Answer:

Last option 3000N

Explanation:

4 0

3 years ago

A basketball with a mass of 0.5 kilograms is accelerated at 2

Paul [167]

Answer: 1N

Explanation: its not 0N.

5 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

A flat uniform circular disk (r= 2.00m,

dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0

3 years ago