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3 years ago

The energy of a photon is 9.7eV what is the wavelength

Physics

1 answer:

grandymaker [24]3 years ago

4 0

Answer:

Wavelength is 0.1278193 mm.

Explanation:

E (photon energy) = h (Planck constant) times c (speed of light in vacuum) divided by λ (photon's wavelength)

h and λ are constants so photon energy E changes in inverse relation to wavelength λ.

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A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w

balu736 [363]

Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.

-- The buoyant force is precisely the missing 30N .

-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

 Weight = (mass) x (gravity)
 185N = (mass) x (9.81 m/s²)
 Mass = (185N) / (9.81 m/s²) = 18.858 kilograms of frewium

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

 30N = (mass of the displaced water) x (9.81 m/s²)

 Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

 Volume of displaced water = 3,058 cm³

Finally, density of the frewium sample = (mass)/(volume)

 Density = (18,858 grams) / (3,058 cm³) = 6.167 gm/cm³ (rounded)

================================================

I'm thinking that this must be the hard way to do it,
because I noticed that

 (weight in air) / (buoyant force) = 185N / 30N = 6.1666...

So apparently . . .

 (density of a sample) / (density of water) =

 (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.

3 0

3 years ago

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

3 years ago

When a cold air mass and a warm air mass meet but are at a standstill, the boundary is called.

Sphinxa [80]

Answer:

Ur answer is Stationary Front

Explanation:

Stationary Front is when a cold air mass and a warm air mass but are at a standstill the boundary is called Stationary Front.

3 0

2 years ago

The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper

-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N $(5,0.02^2)$