Ask question

Ask question

All categories

2 years ago

9

The average threshold of dark-adapted (scotopic) vision is 4.00 10-11 W/m2 at a central wavelength of 500 nm. If light with this

intensity and wavelength enters the eye and the pupil is open to its maximum diameter of 7.20 mm, how many photons per second enter the eye

1 answer:

Studentka2010 [4]2 years ago

8 0

Answer:

E = h f = h c / λ energy of single photon

n E = 4.00E-11 W/m^2 energy required for visibility

n h c / λ = 4.00E-11 number of photons required for visibility (per square meter)

n = 4.00E-11 * 5.00E-7 / (6.63E-34 * 3.00E8)

n = 20.0E-18 / 19.9 E-26 = 1.00E8

100 million photons would be required on 1 m^2 to create visibility

a / A = π * (3.6E-3)^2 / 1 = 4.07E-5 fraction of area available

4.07E-5 * 1.00E8 = 4.07E3 photons required

You might be interested in

A 2 kg ball of clay moving at 35 m/s strikes a 10 kg box initially at rest. What is the velocity of the box after the collision?

Ainat [17]

Answer:

V = 5.83 m/s

Explanation:

Given that,

Mass of a ball of a clay, m = 2 kg

Initial speed of the clay, u = 35 m/s

Mass of a box, m' = 10 kg

Initially, the box was at rest, u' = 0

We need to find the velocity of the box after the collision. Let V be the common speed. Using the conservation of momentum to find it.

$mu+m'u'=(m+m')V\\\\V=\dfrac{mu+m'u'}{(m+m')}\\\\V=\dfrac{2\times 35+10\times 0}{2+10}\\\\V=5.83\ m/s$

So, the velocity of the box after the collision is equal to 5.83 m/s.

4 0

3 years ago

A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a

atroni [7]

Answer:

0.0195 m

Explanation:

$\rho _{p}$ = density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³

$d_{p}$ = diameter of hockey puck = 13 cm = 0.13 m

$h_{p}$ = height of hockey puck = 2.8 cm = 0.028 m

$\rho _{m}$ = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

$d$ = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

$\rho _{m} (0.25)(\pi d_{p}^{2} d ) g = \rho _{p} (0.25)(\pi d_{p}^{2} h_{p} ) g\\\rho _{m} ( d ) = \rho _{p} ( h_{p} )\\(13600) d = (9450) (0.028)\\d = 0.0195 m$

7 0

3 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the <u>centripetal acceleration</u>

$V$ is the<u> tangential speed</u>

$r=12 m$ is the <u>radius</u> of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

<u />

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago

Which statements describe a situation in which work is being done? Check all that apply.

andreev551 [17]

You didn’t submit a photo

5 0

3 years ago

Light travels at a speed of about 3.0 108 m/s. (a) how many miles does a pulse of light travel in a time interval of 0.1 s, whic

babunello [35]

Speed of light is given as

$c = 3 * 10^8 m/s$

time interval is given as

$\Delta t = 0.1 s$

so the distance covered by the light is given as

$d = v * \Delta t$

$d = 3 * 10^8 * 0.1 = 3 * 10^7 meter$

now as we know that

$1 mile = 1609 meter$

so the distance moved by light is

$d = \frac{3 * 10^7}{1609} = 18645.12 miles$

now for the comparision of this distance with diameter of earth

as we know that radius of earth is

$R = 6.38 * 10^6 m$

so the diameter of earth will be

$d = 2R = 12.76 * 10^6 m$

now the ratio of diameter with the distance that light move will be

$\frac{distance}{diameter} = \frac{3 * 10^7}{12.76 * 10^6}$

$\frac{distance}{diameter} = 2.35$

<em>so it is 2.35 times more than the diameter of earth</em>

4 0

3 years ago