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Margaret [11]
3 years ago
6

A trained eye in the dark for an extended period of time may pick up a light stimulus from a light source, at the lowest radiate

d power of 1,2*10* (- 17) W. Determine how many photons of light with a wavelength of 500nm fall on the retina of the eye every second
Physics
1 answer:
frez [133]3 years ago
3 0

Answer:

    #_photons = 30 photons / s

Explanation:

Let's start by finding the energy of a photon of light, let's use the Planck relation

         E = h f

the speed of light is related to wavelength and frequency

         c = λ f

we substitute

         E = h c /λ

         E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹

         E₀ = 3.978 10⁻¹⁹ J

now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second

          #_photons = 1 photon   (E / Eo)

          #_photons = 1  1.2 10⁻¹⁷ /3.978 10⁻¹⁹

          #_photons = 3.0 10¹

          #_photons = 30 photons / s

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Explanation:

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Substances can undergo physical changes or chemical changes. what is the difference between these two kinds of changes
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3 years ago
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In the development of atomic models, it was realized that the atom is mostly empty. Consider a model for the hydrogen atom where
Rama09 [41]

Answer:

Check the explanation

Explanation:

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8 0
3 years ago
You are given aqueous solutions of six different substances and asked to determine whether they are strong, weak, or nonelectrol
kogti [31]

Answer:

Answer is explained below;

Explanation:

Electrolytes are any substances that dissociate into charged particles called ions when dissolved in water. The positively charged ions called cations and the negatively charged ions called anions move toward the negative and positive terminals (cathode and anode) of an electric circuit.

When a substance dissolved in water completely dissociates into ions, it is called a strong electrolyte. The aqueous solutions containing strong electrolytes conduct electricity very well and the examples include strong acids and soluble ionic compounds such as barium chloride, sodium hydroxide, etc.  

When a substance dissolved in water does not completely dissociate into ions, it is called a weak electrolyte. Since the aqueous solutions containing weak electrolytes have relatively few ions, their electrical conductivity is very low compared to the solutions containing strong electrolytes. Examples of weak electrolytes include weak acids and bases like acetic acid, ammonia, etc.

When a substance does not dissociate into ions when dissolved in water, it is called a nonelectrolyte. Since the aqueous solutions containing nonelectrolytes do not contain any ions, such solutions do not conduct electricity. Examples of nonelectrolytes are ethanol, aldehydes, glucose, ketones, etc.

If a solution contains dissolved ions, it conducts electricity and as the ion concentration increases, the conductivity also increases. To determine whether the aqueous solutions of six different substances are strong, weak, or nonelectrolytes, we can test them by applying a voltage to electrodes immersed in the solutions and a light bulb. By observing the brightness of the light bulb or by measuring the flow of electrical current, we can find out which solution contains a strong electrolyte or weak electrolyte, or nonelectrolyte.

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3 0
3 years ago
A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele
artcher [175]

Answer:

<em>the net work done after starting from rest is =  2.4 × 10⁵ J</em>

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J),  The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

<em>Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)</em>

<em>Substituting these values into equation 2</em>

<em>v² = 0² +2×1×30</em>

<em>v² = 60</em>

<em>v = √60</em>

<em>v = 7.75 m/s</em>

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

<em>Substituting these values into equation 1,</em>

<em>W = 1/2(8.0×10³)(7.75)²</em>

<em>W = (4.0×10³)(60)</em>

W = 240 × 10³ J

<em>W = 2.4 × 10⁵ J</em>

<em>Therefore the net work done after starting from rest is =  2.4 × 10⁵ J</em>

4 0
3 years ago
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