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astraxan [27]
3 years ago
10

What is the longest wavelength of radiation that possesses the necessary energy to break the o-o bond?

Physics
1 answer:
Vika [28.1K]3 years ago
7 0

Answer:

The longest wavelength of radiation is 241nm and it lies in ultraviolet region.

Explanation:

The minimum energy required to break o-o bond is 495kJ/mole.

The photon does not have mass and the energy of the single photon depends entirely on the wavelength and is given by

e =hc/λ

where, h is the Planck constant,

            c is the speed of light

            e is the energy of photon

            λ is the wavelength

From the Planck formula we can understand that energy of the photon is quantized.

E = e.Nₐ

e = E/Nₐ =\frac{495KJ}{6.022*10^{23} } = 8.22*10⁻¹⁹ J/photon

λ = hc/E= \frac{6.626*10^{-34}*2.998*10^{8}  }{8.22*10^{-19} }

λ = 2.41*10⁻⁷ m = 241 nm

The longest wavelength of radiation is 241nm and it lies in ultraviolet region.

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Answer:

A - 0 N

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Which statement is not a good practice when working inside a computer case?
lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

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3 years ago
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

2gh = v_f^2 - v_i^2

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\

<u>h = 0.82 m</u>

Now, for the time in air during upward motion we use first equation of motion:

v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s

(c)

Now we will consider the downward motion and use the third equation of motion:

2gh = v_f^2-v_i^2

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\

<u>vf = 7.17 m/s</u>

Now, for the time in air during downward motion we use the first equation of motion:

v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

<u>t = 1.14 s</u>

7 0
2 years ago
Why do astronauts in orbit in space shuttle seem to float
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No gravity in space.

4 0
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If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?
Illusion [34]

The gravitational potential energy U is defined as the product of mass m, the acceleration of gravity g and the height of object h.

U = mgh

We do not have the mass of the hiker. But we know that its W weight is:

W = mg

Where

g = 9.8 m/s^2

So:

m = \frac{W}{g}\\\\m = \frac{600}{g}.

So:

U = (\frac{600}{g})gh\\\\U = (600N)(50m)

U = 30000J

The hiker has gained 30,000 J of energy

6 0
3 years ago
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