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hammer [34]
3 years ago
8

A discharge lamp rated at 25 W (1 W = 1 J/s) emits yellow light of wavelength 580 nm. How many photons of yellow light does the

lamp generate in 1.0 s?
Physics
1 answer:
Scrat [10]3 years ago
3 0

Answer:

the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 10^{19}

Explanation:

given information:

power, P = 25 W

wavelength. λ - 580 nm = 5.80 x 10^{-7} m

time, t = 1 s

to calculate the number of photon(N), we use the following equation

N = λPt/hc

where

λ = wavelength (m)

P = power (W)

t = time interval (s)

h = Planck's constant (6.23 x 10^{-34} Js)

c = light's velocity (3 x 10^{8} m/s^{2})

So,

N = λPt/hc

   = (5.80 x 10^{-7})(25)(1)/(6.23 x 10^{-34})(3 x 10^{8} m/s^{2})

    = 7 x 10^{19}

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A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
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Answer:

v = 5.34[m/s]

Explanation:

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E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

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In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

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E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

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4 0
3 years ago
The SI unit of average spped and velocity.​
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2 years ago
A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is
GREYUIT [131]

Answer:

\omega = 22.67 rad/s

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

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now we know that work done is product of force and displacement

so here we have

W = F.d

W = (44 N)(0.9 m) = 39.6 J

now for moment of inertia of the disc we will have

I = \frac{1}{2}mR^2

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now from above equation we will have

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5 0
3 years ago
Read 2 more answers
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max2010maxim [7]

Answer:

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Explanation:

Given:

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Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

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3 years ago
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