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Stella [2.4K]
4 years ago
10

Ultraviolet (UV) radiation can be subdivided into three regions: UVA, UVB, and UVC, based on their energy. What are the minimum

and maximum values of wavelength in the UVC region?
Physics
1 answer:
ruslelena [56]4 years ago
4 0

Answer:

100nm-280nm

Explanation:

Ultraviolet rays (UV) are part of the electromagnetic spectrum. It goes from 10nm to 400nm wavelengths, they are shorter than visible light, thus it's impossible to see by a human eye, and larger than X-rays (used in many medical applications and harmful when long-exposed).

According to its wavelengths, UV can be divided in different types:

UVA: long wave UV (315nm-400nm)

UVB: medium-wave UV (280nm-315nm)

UVC: short wave UV (100nm-280nm)

Therefore, UVC comprises wavelengths between 10nm and 280nm.

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A steel ball moves from a position of +125 meters to a position of -75 meters. This motion takes 90.0 seconds. What is the veloc
bonufazy [111]

Answer:

2.22m/s to the left

Explanation:

Given parameters:

Initial position = +125m

Final position  = -75m

Motion time  = 90s

Unknown:

Velocity of the steel ball  = ?

Solution:

The velocity of the steel ball is given as the displacement divided by the time;

   Velocity  = \frac{displacement}{time}

The net displacement of the ball  = 125- (-75) = 200m to the left

Input the parameters and solve for the velocity;

     Velocity  = \frac{200}{90}   = 2.22m/s to the left

3 0
4 years ago
Two 2.0cm×2.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery. What is the
Fantom [35]

Answer:

The charge on each electrode is

q=0.354 x10^{-12}

Explanation:

The charge to find the equation is development

V=9v

d=1.0 mm\\A=2cm*2cm=0.02m*0.02m\\A=4x10^{-4} \\E_{o} =8.85x10^{-12} (\frac{C^{2} }{N*m^{2} } )

q=C*V

C=\frac{E_{o}*A }{d} \\C=\frac{8.85x10^{-12} *4x10^{-4}  }{0.01}=0.354  \\

q=0.354*9v\\q=3.186x10^{-12} C

3 0
4 years ago
. Suppose an electron is chasing a proton up this page when suddenly a magnetic field pointing into the page is applied. What wo
irina1246 [14]

The proton and electron experience forces to the left and right of the page, the particles would separate to the left and right respectively after the magnetic field is applied.

To know what would happen to each particle when a magnetic field pointing in the direction of the page is applied, we need to know the direction of the force acting on the particles due to this magnetic field according to fleming's left hand rule

<h3>What is Fleming's left hand rule?</h3>

This states that when the thumb, the middle-finger and fore-finger of the left-hand are held at right angles to each other, the magnetic field points in the direction of the fore-finger, the current (or direction of charge) points in the direction of the middle finger and the force on the conductor or charge points in the direction of the thumb.

<h3>What is the direction of force on the proton?</h3>

Since the direction of the proton is up the page and the direction of the magnetic field is into the page, the direction of the force according to fleming's left hand rule is to the left of the page.

<h3>What is the direction of force on the electron?</h3>

Since the direction of the electron is up the page (it is in the direction of the conventional current or positive charge in the other direction) and the direction of the magnetic field is into the page, the direction of the force according to fleming's left hand rule is to the right of the page.

Since both the proton and electron experience forces to the left and right of the page, the particles would separate to the left and right respectively after the magnetic field is applied.

Learn more about magnetic field here:

brainly.com/question/26051825

8 0
2 years ago
49. In what direction does the force of friction act?
DedPeter [7]
I believe it always acts in the direction opposite to that of the object in motion, so if the force was applied to the left then friction acts in the right

hope that helps
5 0
3 years ago
If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th
alex41 [277]
Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.  
So we have 1/2 MV^2 = MGH 
V^2 = 2GH 
V = âš2GH 
V = âš( 2 * 9.8 * 325)  
V = âš 6370
 V = 79.81 m/s
6 0
3 years ago
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