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BigorU [14]
3 years ago
15

Viewed through a spectroscope, the spectral profile of a yellow street lamp has a narrow line in the yellow region of the visibl

e spectrum against a black background. Is the street lamp an incandescent or an atomic source of light
Physics
1 answer:
muminat3 years ago
6 0

Answer:

discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source

Explanation:

Bulbs can emit light in several ways:

* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.

* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.

Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type

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A 60-W light bulb runs on 120 V and draws 0.50 A of current when running
Pavlova-9 [17]

Answer:

3600 J

Explanation:

According to given question

P(rated)=60w

V= 120

I =0.50 A

t=600 second

Now,

Energy can be calculated as :

E= VI*t

Where,

V is voltage

I is current

t is time in second

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Putting the all value in above equation E

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E=120*0.5*600\\E=36000 J

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8 0
3 years ago
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the
Elina [12.6K]

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

          T = 2\pi  \sqrt{ \frac{L}{g} }

          T² =  4\pi ^2 \frac{L}{g}4pi2 L / g

          g = 4\pi ^2   \frac{L}{T^2}

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

           g = 4\pi ^2  \frac{0.823}{1.823^2}4 pi2 0.823 / 1.823 2

            g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

             T = t / n

             ΔT = \frac{dT}{dt} Δt + \frac{dT}{dn} ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

              ΔT = Δt / n

              ΔT = Δt

now let's look for the uncertainty of g

             Δg = \frac{dg}{dL} ΔL + \frac{dg}{dT}  ΔT

             Δg = 4\pi ^2 \frac{1}{T2}   ΔL + 4π²L  (-2  T⁻³) ΔT

           

a more manageable way is with the relative error

             \frac{\Delta g}{g}   = \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}

we substitute

              Δg = g ( \frac{\Delta L }{L} + \frac{1}{2}  \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation  

let's calculate

               Δg = 9.7766 (\frac{0.001}{0.823} + \frac{1}{2}  \ \frac{0.671}{1.823})

               Δg = 9.7766 (0.001215 + 0.0184)

               Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

                Δg = 0.2 m / s2

therefore the correct result is

               g ±Δg = (9.8 ± 0.2) m / s²

5 0
3 years ago
Please solve the Problem.
babymother [125]

From the graph, the average acceleration between 5.0 s and 8.0 s is 2.33 ms^2.

<h3>What is acceleration?</h3>

The term acceleration has to do  with the change in velocity with time. We can see that the graph shown is a graph of velocity against time, the slope of the graph is the acceleration.

Thus, the average acceleration between 5.0 s and 8.0 s can be read off from the graph as 20 m/s - 13m//8.0 s - 5.0 s = 2.33 ms^2.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

8 0
2 years ago
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