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BigorU [14]
2 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:
muminat2 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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Read the scenario and solve these two problems.
Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy K of an object is given by:

K=\frac{1}{2}mV^{2}

Where:

m=12000 kg is the mass of the train

V=30 m/s is the speed of the train

Solving the equation:

K=\frac{1}{2}(12000 kg)(30 m/s)^{2}

K=5400000 J This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

E_{i}=E_{f}

K_{i}+P_{i}=K_{f}+P_{f}

Where:

K_{i}=5400000 J is the train's initial kinetic energy

P_{i}=0 J is the train's initial potential energy

K_{f}=0 J is the train's final kinetic energy

P_{f}=mgh is the train's final potential energy, where g=9.8 m/s^{2} is the acceleration due gravity and h is the height.

Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

Finding h:

h=45.918 m \approx 45.92 m

7 0
3 years ago
Read 2 more answers
An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0
3 years ago
Read 2 more answers
Three balls with different masses are shown below.
algol [13]

Answer:

$10 a g

Explanation:

6 0
3 years ago
Read 2 more answers
What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?
Alexandra [31]
600/3 = 200
the slope is 200m/min
 
OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour 

6 0
2 years ago
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