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Gelneren [198K]
4 years ago
7

A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second

Physics
2 answers:
hjlf4 years ago
6 0
The power that the light is able to utilize out of the supply is only 0.089 of the given.
                           Power utilized = (0.089)(22 W)
                                                  = 1.958 W
                                                  = 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
                           Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J. 
Charra [1.4K]4 years ago
4 0

On Apex the answer is 1.96J



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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
A car of mass 800kg travels a distance of 40m at constant speed in a duration of 2.0s. The car exerts a forward force of 15kN.
Alex17521 [72]

W = F × s

W = 15kN × 40 m

W = 15.000 N × 40 m

W = 600.000 J

P = W/t

P = 600.000 J/2 s

P = 300.000 Watt

P = 300kWatt

#LearnWithEXO

6 0
3 years ago
When turning a magnet very quickly would be BEST used to create A) radiation. B) light waves. C) an electric current. D) a conve
Gre4nikov [31]

the correct choice is

C) an electric current.

as a magnet is turned quickly relative to a coil, the magnetic flux linked with coil varies due to variation of angle of direction of magnetic field with normal to the plane of coil. the coil resist this change of magnetic flux in it by inducing emf in it so as to nullify the variation in magnetic flux. Due to this induced emf , electric current flows through the coil.

6 0
3 years ago
Read 2 more answers
Question 2 of 10
Mekhanik [1.2K]

the awnser to ur question is D

6 0
3 years ago
How can a magnetic field be produced, using a wire, a battery, and and a nail?
Aleks [24]
It is fairly easy to build an electromagnet. All you need to do is wrap some insulated copper wire around an iron core. If you attach a battery to the wire, an electric current will begin to flow and the iron core will become magnetized. When the battery is disconnected, the iron core will lose its magnetism. Follow these steps.
Step 1 - Gather the Materials
One iron nail fifteen centimeters (6 in) long
Three meters (10 ft) of 22 gauge insulated, stranded copper wire
One or more D-cell batteries
Step 2 - Remove some Insulation
Step 3 - Wrap the Wire Around the Nail
Step 4 - Connect the Battery
5 0
3 years ago
Read 2 more answers
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