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2 years ago

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

r atmosphere is ozone, O3O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 HzHz . What is the wavelength λλlambda of the radiation absorbed by ozone? Express your answer in nanometers.
λ = ___.

2 answers:

vesna_86 [32]2 years ago

5 0

Answer:

0.03052 nm

Explanation:

From c = f¥

Where c = speed of electromagnetic wave in vacuum = 3x10^8 m/s

f = frequency of the wave

¥ = wavelenght of the wave.

¥ = c/f = (3x10^8)/(9.38×10^14)

= 3.052x10^-7 m

= 0.03052 nm

irakobra [83]2 years ago

3 0

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>

So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.

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A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

irina [24]

Answer:

The atomic weight in g/mole of the metal (molar mass) is 8.87.

Explanation:

To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The chemical formula of metal oxide tells you that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

In the previous equation you can see that you need 3 oxygen anions to react with two metal cations. Then:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

You have 52.92 g of metal in the sample, then the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>

The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0

2 years ago

Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the

enyata [817]

<u>Answer</u>

27.7

<u>Explanation</u>

The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.

cosx = adjacent/hypotenuse

cos 30 = adjacent / 10

adjacent = 10 cos30

= 8.66 m/s ⇒ This is the horizontal speed.

Now find the horizontal distance.

Distance = speed × time

= 8.66 × 3.2

= 27.71

Answer to the nearest tenth = 27.7

4 0

2 years ago

Read 2 more answers

You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg

Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball $m=0.0570\ kg$

ball leaves the hand with $u=15\ m/s$

maximum height reached by ball $h=8\ m$

Initial Mechanical energy when ball just leaves the hand

$M.E._1=(P.E.+K.E.)_1$

$M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1$

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

$M.E._1=\frac{1}{2}\times m\times (15)^2$

$M.E._1=6.41\ J$

Mechanical Energy at highest point

$(M.E.)_2=(P.E.+K.E.)_2$

at highest Point velocity is zero

$(M.E.)_2=mgh_2+0$

$(M.E.)_2=0.0570\times 9.8\times 8$

$(M.E.)_2=4.46\ J$

Decrease in Mechanical energy

$(M.E.)_1-(M.E.)_2=6.41-4.46$

$(M.E.)_1-(M.E.)_2=1.95\ J$

3 0

2 years ago

Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b

Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

$\Delta ACB$

$tan 30 = \frac{8.3}{h}$

$h= \frac{8.3}{tan 30} = 14.376 m$

therefore

height of keisha is $= h + 1.5 m$

= 14.376 + 1.5

$= 15.876 \simeq 16 m$

therefore option c is correct

5 0

2 years ago

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th

jek_recluse [69]

Answer:

$a)v_{f}=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}$ (1)

At the beginning the girl is stationary:

$v_{o.girl}=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_{f.girl}=v_{f.ball}=v_{f}$ (3)

We replace (2) and (3) in (1):

$m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\$

We can now solve the equation for v_{f}:

$v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s$

4 0

2 years ago