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algol [13]
3 years ago
5

The wavelengths of light emitted by a firefly span the visible spectrum but have maximum intensity near 550 nm. A typical flash

lasts for 100 ms and has a power of 1.2 mW. If we assume that all of the light is emitted at the peak-intensity wavelength of 550 nm, how many photons are emitted in one flash?
Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

3.3\cdot 10^{15}

Explanation:

First of all, let's calculate the energy of a single photon of wavelength

\lambda=550 nm=5.5\cdot 10^{-7}m

which is given by

E_1 = \frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.5\cdot 10^{-7} m}=3.6\cdot 10^{-19} J

The power of the flash is

P=1.2 mW=0.0012 W

and the time it lasts is

t=100 ms=0.1 s

so the total energy delivered in one flash is

E=Pt=(0.0012 W)(0.1 s)=1.2\cdot 10^{-3}J

This energy contains exactly N photons each of energy E_1, so the number of photons emitted in one flash is

N=\frac{E}{E_1}=\frac{1.2\cdot 10^{-3} W}{3.6\cdot 10^{-19}J}=3.3\cdot 10^{15}

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5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How
ad-work [718]

Answer:

60 m/s

Explanation:

a =  \frac{vf - vi}{t}

5.0 =  \frac{vf - 0}{12}

5.0 \times 12 = vf

60 = vf

6 0
3 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

         F - fr = 0

         F = fr

         fr = 52.1 N

Now we can work  with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

        N - W₁ -W₂ = 0

        N = W₁ + W₂

as the two blocks are identical

        N = 2W

X axis

        F - fr₁ - fr₂ = 0

        F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

          fr₁ = 52.1 N

          fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

          fr₂ = μ 2N

          fr₂ = 2 μ N

          fr₂ = 2 fr₁

we substitute

          F = fr₁ + 2 fr₁

          F = 3 fr₁

          F = 3  52.1

          F = 156.3 N

7 0
2 years ago
During science class, while studying mixtures, you mix together iron filings and sand. Your teacher challenges you to separate t
Lapatulllka [165]

Answer : Use a magnet to pull out the iron filings as they are attracted to a magnet

Explanation :  we use the magnet to separate the sand from the iron filing. Because the magnet has an attraction power.

We can say that we can use the magnet to attract the iron filings out of the mixture because iron is magnetic solid, but sand will not attract because sand is not magnetic solid.

So,  we use a magnet to pull out the iron filings as they are attracted to a magnet.

8 0
3 years ago
Read 2 more answers
Forces cause changes to the motion of objects. Name a force and describe two changes it makes.
otez555 [7]
Sliding and Static.

Would be the right one here.
3 0
3 years ago
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