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3 years ago

What is the energy in joules of a mole of photons associated with red light of wavelength 7.00 × 102 nm?

1 answer:

4 0

The energy of a single photon is given by E = hc/lambda, where h is Planck's constant, c is the speed of light, and lambda is the wavelength. Plugging the values in gives E = 6.63E-34 x 3.00E8 / 700E-9 = 2.84E-19 Joules Now one mole of substance is equivalent to 6.02E23 particles, so one mole of these photons will be: 2.84E-19 x 6.02E23 = 1.71E5 Joules

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Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$

dividing both sides by d

$0=-m1g+m2g+\frac{kd}{2}$

$g(m1-m2)= \frac{kd}{2}$

$d=\frac{2g(m1-m2)}{k}$ , with k the constant of the spring and g the gravitational acceleration.

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3 years ago

Please solve no.g Anyone!???

steposvetlana [31]

Answer:

The answer is already given... it's 31.25 cm

5 0

2 years ago

In ionic bonding, atoms

Genrish500 [490]

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3 0

3 years ago

Why would you expect the speed of light to be slightly less in the atmosphere then in a vacuum?

azamat

The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material

Explanation:

When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.

But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.

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3 years ago

Why does a round pizza come in a square box? Random questions

denpristay [2]

Answer:

because they dont know how big it willactually be? idk

or so deep dish can fit too

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3 years ago

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