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Nesterboy [21]
3 years ago
5

A typical 100-watt light bulb consumes 100 j of energy per second. If a light bulb converts all of this energy to 400 nm light

Physics
1 answer:
VladimirAG [237]3 years ago
3 0

The missing question is: how many photons are produced per second?

Answer:

2.0\cdot 10^{20}

Explanation:

The energy of a photon is given by

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

In this problem, the photons have wavelength

\lambda = 400 nm = 400 \cdot 10^{-9} m, so each photon has an energy of

E_1 = \frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{400\cdot 10^{-9}}=4.97\cdot 10^{-19} J

The total energy emitted by the bulb in 1 second is

E = 100 J

Therefore, the number of photons emitted per second is

n=\frac{E}{E_1}=\frac{100}{4.97\cdot 10^{-19}}=2.0\cdot 10^{20}

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Just as a skydiver steps out of the helicopter with no forward velocity someone who’s watching start stopwatch so the time is ze
Kobotan [32]

Answer: zero.


Justification:


The downward velocity of the sky diver just before starting to fall is zero, assuming that the helicopter is not moving but just hovering.


Before starting to fall, the velocity of the skydiver is the same of the helicopter, which is zero. It is only, once she jumps out of the helicopter that her weight is not supported by the helicopter and so the gravitational force of the Earth attracts the skydiver and she starts to gain velocity at a rate equal to g (around 9.81 m/s²).

6 0
3 years ago
Steam enters a long, horizontal pipe with an inlet diameter of D1 = 16 cm at 2 MPa and 300°C with a velocity of 2.5 m/s. Farther
polet [3.4K]

Answer:

m = 0.4005 kg/s

Q_out = 45.1 KJ/s

Explanation:

Given

Pipe inlet diameter D1 = 16 cm

Steam inlet pressure P1 = 2 Mpa

Steam inlet temperature T1 = 300 °C

Pipe outlet diameter D2 = 14 cm

Steam inlet velocity V1 = 2.5 m/s

Steam outlet pressure P2 = 1.8 MPa

Steam outlet temperature T2 = 250 °C  

Required

Determine

(a) The mass flow rate of steam.

(b) The rate of heat transfer.  

Assumptions

Kinetic and potential energy changes are negligible.

This is a steady flow process.

There is no work interaction.  

Solution

Part a From steam table (A-6) at P1 = 2 Mpa , T1 = 300 °C

vl = 0.12551 m^3/Kg

h1 = 3024.2 KJ/Kg

The mass flow rate of steam could be defined as the following  

m = 1/v1*A1*V1

m = 0.4005 kg/s

Part b We take the pipe as our system.The energy balance could be defined as the following  

E_in -E_out =ΔE_sys = 0

E_in = E_out

mh_1 = Q_out + mh_2

Q_out = m(h_1-h_2)

From steam table (A-6) at P2= 1.8 Mpa T2 = 250 °C  h2= 2911.7 KJ/Kg The heat transfer could  be defined as the following

Q_out = m(h_1-h_2)

Q_out = 0.4005*(3024.2 -2911.7) =45.1 KJ/s

6 0
3 years ago
Three resistors are connected into the section of a circuit described by the diagram. At which labeled point or points of the ci
tangare [24]

Answer:

Pont z only

Explanation:

5 0
3 years ago
Fernando vai à pé pra escola. Todos os dias ele sai as 6:30 e chega na escola as 6:55. Nisso ele percorre 2km. Na volta pra casa
alexgriva [62]
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5 0
3 years ago
You pull a rope to the left with 300n and a friend pulls it the rope to the right with 425n what is the magnitude and direction
eduard

Answer: from vector law opposite we minus

Hence

425n-300n=125n to the right

Hints in left vector have -Ve sign while right have +Ve sign

5 0
3 years ago
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