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3 years ago

Find the wavelength (in nm) of a 2.52 eV photon. a. 492.34127 nm.b. 585.88611 nm.c. 541.5754 nm.d. 418.49008 nm.

Physics

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

Answer:

The correct option is (a).

Explanation:

Given that,

The energy of photon, E = 2.52 eV

We need to find the wavelength of the photon in nm. The formula for the energy of a photon is given by :

$E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}\\\\\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{2.52\times 1.6\times 10^{-19}}\\\\=4.93\times 10^{-7}\ m\\\\=493\ nm$

The nearest option is a) i.e. 492.34127 nm.

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A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago

A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th

Evgen [1.6K]

Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

θ = 64.53°

Now by putting the value in equation

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Therefore rate of change in the angle is 0.038\ rad/s

8 0

3 years ago

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3

vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

$k = 5\:\text{N/m}$

The spring potential energy $PE_s$ is

$PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2$

$\:\:\:\:\:\:\:=22.5\:\text{J}$

3 0

3 years ago

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

frutty [35]

Answer:

The distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

We know that the conversion of 1 millimeters is equal to $10^{-3}$ meter. And then the conversion of 1 meter is equal to $10^{-3}$ km. Then the conversion of 1 millimeter to km will be

1 mm = $10^{-3}$ m

1 m = $10^{-3}$ km

So, 1 mm = $10^{-3}$ × $10^{-3}$ km = $10^{-6}$ km.

As here the the distance is 4012 mm, then the distance in km will be

4012 mm = 4012 × $10^{-6}$ km.

So the distance is 4012 × $10^{-6}$ km.

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3 years ago

Which of the following best describes the human population from early times to the present?

Makovka662 [10]

Answer:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

Explanation:

From the figure attached we see the evolution of the human population since early times (1050).

We see that from 1050 until 1750-1850 we have an increase slowly with a low value for the increase per year.

But after these years (1750-1850) we see a considerable increase of the population, like an exponential model.

So then we can conclude in general terms this:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

7 0

3 years ago