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3 years ago

What type of energy can take up space

Physics

1 answer:

Arada [10]3 years ago

8 0

Answer:

Yes energy does take up space.

Explanation:

Every form of energy has a defining characteristic; sound is the vibration of molecules, electricity is the movement of electrons, and mass is the thing that take up space.

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In a compression wave, particles in the medium move

nikdorinn [45]

in the same direction as the wave

Explanation:

In a compression wave, the particles in the medium moves in the same direction as the wave source.

A wave is generally defined as a disturbance that transmits energy.

There are two types of waves based on the direction through which they are propagated.
Transverse waves are directed perpendicularly in the direction of propagation.
Examples are electromagnetic waves.
Longitudinal waves are parallel to their source. Examples are sound waves, p-waves.
They are made up of series of rarefaction and compression.

learn more:

Waves brainly.com/question/3183125

#learnwithBrainly

8 0

3 years ago

A light with a second-order bright band forms a diffraction angle of 30. 0°. The diffraction grating has 250. 0 lines per mm. Wh

Luden [163]

The distance between two successive troughs or crests is known as the wavelength. The wavelength of the light will be 1000 nm.

How do you define wavelength?

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

Diffraction angle= 30⁰

Diffraction grating per mm= 250

wavelength = ?

Mathematically the equation of bright band is given by

$\rm \lambda= \frac{sin\theta}{nN}$

$\rm \lambda= \frac{sin23^0}{250\times 2}$

$\rm \lambda= 0.000001$ m

$\rm \lambda= 1000 nm$

Hence the wavelength of the light will be 1000 nm.

To learn more about the wavelength refer to the link;

brainly.com/question/7143261

8 0

2 years ago

Read 2 more answers

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

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

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$