All categories

3 years ago

Why are sunlight and gravity not considered matter?

2 answers:

6 0

Sunlight and gravity are not considered matter because sunlight does not have a rest mass that is characteristic of matter while gravity is just a vector component of matter.

<u>Explanation:</u>

Sunlight has energy and momentum that makes it similar to matter but it cannot be considered matter when you consider all the factors that are characteristic of matter. Light is an electromagnetic wave that is made up of photons and the protons do not have rest mass.

Gravity is a vector component of matter that has both magnitude as well as direction and the gravity increases with increase in mass of an object.

Nutka1998 [239]3 years ago

5 0

Answer:

Sunlight and gravity are not considered matter because they don't have mass or take up space.

Explanation:

Gravity is a basic and innate element of matter. If there is no matter then there won't be gravity.

But gravity is not considered to be matter because it is actually a force that attracts the matter. So basically gravity is a sort of force affecting the matter.

Sunlight has so many properties to be called as a matter like it has energy and momentum. But it consists of photons which do not have any rest mass, hence sunlight is not considered as matter.

You might be interested in

I need help guyssss plssss

Kipish [7]

Answer:

Cd(NO3)2 + Na2S --> CdS + 2 NaNO3

Explanation:

8 0

3 years ago

The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?

Svetradugi [14.3K]

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

3 0

3 years ago

Which statements describe magnetic poles? Check all that apply.

Alex_Xolod [135]

Explanation:

Magnet: It has two poles: South pole and North pole.

Magnetic field lines are stronger near the poles of the magnet.

Same poles repel each other. There is a magnetic force of repulsion between the same poles. North- North poles repel each other.

Unlike poles attract each other. There is magnetic force of attraction between the opposite poles. South- North poles attract each other.

Mono poles cannot exist.

From the given statements, the magnetic poles are described by:

A north pole must exist with a south pole.

Two south poles placed near each other will repel each other.

A north pole and a south pole placed near each other will attract each other.

5 0

2 years ago

Read 2 more answers

0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

3 years ago

What are the two applications of electron beam welding

Georgia [21]

First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.

Please visit out website at the URL below and you can click the "View Application" button under each listed Industry segment to view case studies of commonly EB welded applications.

https://www.ptreb.com/electron-beam-welding-applications

And for more general information on our welding process, we have an informational section you can peruse as well:

https://www.ptreb.com/electron-beam-welding-information

Good luck with your assignment- we are glad to hear they are teaching about EBW in high school!!!

5 0

3 years ago