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3 years ago

A visible light has a wavelength of 727.3 nm. Determine its frequency, energy per photon, and color.

Physics

1 answer:

Snezhnost [94]3 years ago

3 0

Answer:

$f=4.12\times 10^{14}\ Hz$ and $E=2.73\times 10^{-19}\ J$

Explanation:

The wavelength of a visible light is 727.3 nm.

$727.3\ nm=727.3 \times 10^{-9}\ m$

The formula is as follows :

$c=f\lambda$

f is the frequency of the visible light

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{727.3 \times 10^{-9}}\\\\f=4.12\times 10^{14}\ Hz$

Energy of a photon is given by :

E = hf, h is Planck's constant

$E=6.63\times 10^{-34}\times 4.12\times 10^{14}\\\\E=2.73\times 10^{-19}\ J$

Red color has a frequency of $4.12\times 10^{14}\ Hz$ and energy per photon is $2.73\times 10^{-19}\ J$ .

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Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

What is the SI unit for kinetic energy

trapecia [35]

The easiest way to build a unit for energy is to remember that
'work' is energy, and
   Work = (force) x (distance).

So energy is    (unit of force) x (unit of distance)

   [Energy] =   (Newton) (meter) .

'Newton' itself is a combination of base units, so
energy is really
   (kilogram-meter/sec²) (meter)

   = kilogram-meter² / sec² .

That unit is so complicated that it's been given a special,
shorter name:
   Joule .

It doesn't matter what kind of energy you're talking about.
Kinetic, potential, nuclear, electromagnetic, food, chemical,
muscle, wind, solar, steam ... they all boil down to Joules.

And if you generate, use, transfer, or consume 1 Joule of
energy every second, then we say that the 'power' is '1 watt'.

3 0

3 years ago

Read 2 more answers

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

mario62 [17]

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

4 0

3 years ago

Read 2 more answers

A compass needle acts as a(n)

beks73 [17]

Explanation:

A compass needle acts as a Magnet

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3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago