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Alik [6]
3 years ago
12

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, O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 Hz . What is the wavelength λ lambda of the radiation absorbed by ozone?
Physics
1 answer:
elena55 [62]3 years ago
7 0

Answer:

3.2\times 10^{-7}\ m or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = 9.38\times 10^{14}\ Hz

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = 3\times 10^8\ m/s

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

v=f\lambda

Now, expressing the above equation in terms of wavelength 'λ', we have:

\lambda=\frac{v}{f}

Now, plug in the given values and solve for 'λ'. This gives,

\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m

Therefore, the wavelength of the radiations absorbed by the ozone is nearly 3.2\times 10^{-7}\ m or 0.32 μm.

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A pilot claims to have seen a UFO moving initially at a speed of about 351 m/s in an easterly direction and then, in a time inte
serious [3.7K]

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .

3 0
3 years ago
You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.
Blizzard [7]

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0
3 years ago
What is the kinetic energy of a 7.56 kg bowling ball that is rolling at a speed of 2 m/s ?
postnew [5]

Answer:

K.E=?

Mass=7.56 kg

Velocity=2m/s

Solution::

K.E = 1/2mv^2

K.E=0.5*7.56*(2)^2

K.E= 0.5*7.56*4

K.E=15.12 joules.

7 0
3 years ago
Two forces, F1 and F2, are applied to a block on a frictionless, horizontal surface
Komok [63]

Answer:

F, = 12N. F, = 2 N. Block. 4) a 20.0-kg mass moving at 1.00 m/s.

Explanation:

5 0
2 years ago
Read 2 more answers
A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?
uranmaximum [27]
Simply subtract the two velocities and divide by 8.1,

\frac{0 - 35}{8.1} = - 4.32

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
5 0
3 years ago
Read 2 more answers
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