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Anna007 [38]
3 years ago
10

Dinoflagellates are single-cell creatures that float in the world's oceans; many types are bioluminescent. When disturbed by mot

ion in the water, a typical bioluminescent dinoflagellate emits 100,000,000 photons in a 0.10s long flash of light of wavelength 460 nm .
What is the power of the Flash in watts?
Physics
1 answer:
arlik [135]3 years ago
3 0

Answer:

P = 43.2 10⁻¹⁰ W

Explanation:

To find the power, let's start by calculating the energy of each photon using the Planck equation

     E = h f

Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and f is the emitted frequency

The speed of light is related to wavelength and frequency

      c = lam f

Let's replace

    E = h c / lam

    E = 6.63 10⁻³⁴ 3. 10⁸/460 10⁻⁹

    E = 4.32 10⁻¹⁹ J

This is the energy of each photon emitted, the total energy is the product of this value by the number cded photons

Let's calculate the power

     P = W / t = Total E / t

     P= N E /t

    P = 1 10⁹ 4.32 10⁻¹⁹ / 0.10

    P = 43.2 10⁻¹⁰ W

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Me ajudem, Por favor!!!!!!
zzz [600]

Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

d)  Displacement = 22 m

e)  Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

3 0
4 years ago
The gas in a cylinder has a volume of 5 liters at a pressure of 115 kPa. The pressure of the gas is increased to 211 kPa. Assumi
stira [4]
We know the initial pressure and volume, the final pressure, and the temperature is constant, Using the G.U.E.S.S. method we can express the problem like this:                                             
Givens: V_1=5 L, P_1= 115kPa, P_2= 211kPa

Unknown: V_2 
Equation:  \frac{V_1P_1}{P_2}=V_2
Substitute: \frac{5L*115kPa}{211kPa}= V_2
Solve: \frac{5L*115kPa}{211kPa}= 2.73 or 2.72L

      In conclusion, your answer is either 2.73 kPa if you round or 2.72 kPa if you don't. Hope you learned from it.

                                                                            
3 0
3 years ago
Who was the first person to describe the earth as a magnet
denis-greek [22]

Answer:

William Gilbert

Explanation:

first described the Earth as a giant dipole magnet 400 years ago. But, as Rod Wilson recounts, he did far more than this.

5 0
3 years ago
Read 2 more answers
Why does japan have so many volcanoes? <br><br> •paragraph
AlladinOne [14]

Answer:

This reason is that Japan is located along the Pacific 'ring of fire' which is an area along the Pacific plate boundaries where there is a lot of volcanic activity

Explanation:

Japan also lies on the edges of several continental and oceanic plates so this is why Japan experiences a lot of earthquakes.

8 0
3 years ago
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

8 0
3 years ago
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