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Taya2010 [7]
3 years ago
5

Ozone in the upper atmosphere absorbs energy in the 210-230 nmnm range of the spectrum. In what region of the electromagnetic sp

ectrum does this radiation occur
Physics
1 answer:
kifflom [539]3 years ago
5 0

Answer:

Ultraviolet

Explanation:

The electromagnetic spectrum is divided into a number of parts. One of which is the ultraviolet region. The ultraviolet region is found between the 100 nm - 400 nm band of the electromagnetic spectrum. From the range we just stated now, we can see for certainty that the given range 210 nm - 230 nm lies inside of the 100 nm c 400 nm of the electromagnetic spectrum, that of which is the ultraviolet region.

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How do l calculate e<br> nergy
aivan3 [116]

The formula for energy of motion is KE = .5 x m x v^2

Ke= Kinetic Energy in Joules

m = Mass in Kilograms

v = Velocity in Meters per Second

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3 years ago
A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr
Ipatiy [6.2K]

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = \frac{9\times 10^9\times3\times10^{-9}}{d^2}

Field due to charge = 4 X 10⁻⁹ C

E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}

These two fields will be equal and opposite to make net field zero

\frac{9\times 10^9\times3\times10^{-9}}{d^2} = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

\frac{3}{d^2} =\frac{4}{(2-d)^2}

\frac{2-d}{d} =\frac{2}{1.732}

d = 0.928

5 0
3 years ago
What is a safe following distance between your automobile and the vehicle in front of you?
Colt1911 [192]

Answer:

Many drivers follow the “three-second rule.” In other words, you should keep three seconds worth of space between your car and the car in front of you in order to maintain a safe following distance.

Thank you and please rate me as brainliest as it will help me to level up

4 0
3 years ago
Cardiorespiratory fitness improves the efficiency of the cardiovascular and the respiratory systems in delivering oxygen to the
algol [13]
Your answer should be T
4 0
3 years ago
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
3 years ago
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