All categories

3 years ago

Which of the following is true about light?

1 answer:

7 0

Light travels in waves AND in bundles called "photons".

It's hard to imagine something that's a wave and also a bundle.
But it turns out that light behaves like both waves and bundles.

If you design an experiment to detect waves, then it responds to light.

And if you design an experiment to detect 'bundles' or particles, then
that one also responds to light.

You might be interested in

What must an object have in order to exert a gravitational field?

AnnZ [28]

Answer:

power

Explanation:

6 0

3 years ago

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

3 years ago

What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?

sergey [27]

The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

100,000 = 600,000a

a = 0.167m/s²

Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

6 0

2 years ago

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0

3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$