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irinina [24]
3 years ago
5

Light of intensity 2.00 W/m2 passes through the pupil of one of your eyes and eventually falls on your retina. The radius of the

pupil is controlled by the iris. In a human eye, the radius of the pupil can vary from about 1.0 mm to about 3.5 mm. The area of the opening can be calculated from pi (3.1416) times the radius squared. If your pupil is open to a radius of 2.5 mm, how much light energy passes through in one minute
Physics
1 answer:
lord [1]3 years ago
3 0

Answer:

Explanation:

Intensity of light, I = 2 W/m²

radius of pupil open, r = 2.5 mm = 2.5 x 10^-3 m

time, t = 1 min = 60 second

Energy passes, E = intensity x area x time

E = 2 x 3.14 x 2.5 x 2.5 x 10^-6 x 60

E = 2.36 x 10^-3 Joule

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kvasek [131]

Answer:

t = 106π / 30*2.1

Explanation:

w_{i} = 1.06*10^{2}    => 106

    => 106 x 2π/60

    => 106/30π

∝ = -2.1 rad/sec²

w_{f} => 0

w_{f} = w_{i} + ∝t

∴ (w_{f} - w_{i}) / ∝ = t

t = 106π / 30*2.1

6 0
3 years ago
How do you calculate acceleration
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A=f/m
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7 0
3 years ago
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Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.
MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>
  • It is given that, the infinite well having a width of 0.050 mm.
  • We have the expression for energy of electron in an infinite well as,

                  E_n=\frac{n^2h^2}{8mL^2}

  • where;

                m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1

  • Thus, the lowest energy of electron in an infinite well is,

                E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

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7 0
1 year ago
At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen
Misha Larkins [42]

Answer:

  1. The specific mechanical energy of the air in the specific location is 40.5 J/kg.
  2. The power generation potential of the wind turbine at such place is of 2290 kW
  3. The actual electric power generation is 687 kW

Explanation:

  1. The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: \frac{1}{2} V^2 where V is the velocity of the air.
  2. The specific kinetic energy would be: \frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}.
  3. The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
  4. To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: \frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}
  5. Then the mass flow is obtain from the volumetric flow times the density of the air: m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}
  6. Then, the Power generation potential is: 40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW
  7. The actual electric power generation is calculated using the definition of efficiency:\eta=\frac{E_P}{E_I}}, where η is the efficiency, E_P is the energy actually produced and, E_I is the energy input. Then solving for the energy produced: E_P=\eta*E_I=0.30*2290kW=687kW
6 0
3 years ago
Give an example of where you see the law of conservation of energy in your life
lesya692 [45]

Answer:

The law of conservation of energy can be seen in these everyday examples of energy transference: Water can produce electricity. Water falls from the sky, converting potential energy to kinetic energy.

The cue ball loses energy because the energy it had has been transferred to the 8 ball, so the cue ball slows down

Explanation:

8 0
3 years ago
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