This is not something that waves do because they need a medium to travel through, while particles do not.
<h3>How light travels in space?</h3>
A light travels without any medium while on the other hand, a medium is required for sound waves to move from oe place to another. Sound is a mechanical wave that cannot travel through a vacuum.
So we can conclude that electromagnetic waves like light do not require medium for its propagation.
Learn more about light here: brainly.com/question/19697218
Conditions for equilibrium require that the sum of all external forces acting on the body is zero (first condition of equilibrium), and the sum of all external torques from external forces is zero (second condition of equilibrium). These two conditions must be simultaneously satisfied in equilibrium.
Answer:
Distance: -30.0 cm; image is virtual, upright, enlarged
Explanation:
We can find the distance of the image using the lens equation:

where:
f = 15.0 cm is the focal length of the lens (positive for a converging lens)
p = 10.0 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q,

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).
The magnification can be found as

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.
Also, the fact that the magnification is positive tells us that the image is upright.
Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.
You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.
The parallel combination of the first two resistors looks like a single
resistor, whose value is
1 / (1/6 + 1/12)
= 1 / (2/12 + 1/12)
= 1 / (3/12)
= 12/3 = 4Ω .
Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of
4Ω + 2Ω = 6Ω .
So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is
I = V / R = (24v) / (6Ω) = 4 Amperes.
That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.
So the current through the 2Ω resistor is 4 Amperes. (B).
Note:
The POWER dissipated by the 2Ω resistor is
P = I² R = (4A)² · (2Ω) = 32 watts .
This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.
Remember the following formulas to convert temperatures between Farenheit and Celsius:

Convert all the temperatures to Celsius to compare them.
<h2>
Part a)</h2>
Convert 265ºF to Celsius:

Then, 265ºF is equal to 129.444...ºC. Then, 265ºC is higher than 265ºF.
The correct choice is option 2) 265ºC.
Part b)
Convert 370ºF to Celsius:

Then, 370ºF is equal to 187.777...ºC. Then, 200ºC is higher than 370ºF.
The correct choice is option 2) 200ºC.