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Katena32 [7]
2 years ago
15

Determine the minimum number of photons that must enter the eye's pupil each second in order for an object to be seen. Assume th

at the pupil's radius is 0.20 cm and the wavelength of the light is 550 nm . To see an object with the unaided eye, the light intensity coming to the eye must be about 5×10−12J/m2⋅s or greater. Calculate Nmin... I keep getting ​1.38 e25 which is not right and I can figure out what I am doing wrong.
Physics
1 answer:
irina1246 [14]2 years ago
8 0

Answer:

Explanation:

Energy of a photon = hν , h is plank's constant and ν is frequency .

=    h c / λ     ; c is velocity of light and  λ is wavelength .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 550 x 10⁻⁹

= .036 x 10⁻¹⁷ J  

Let no of photons be n .

Total energy of n photons

= n x  .036 x 10⁻¹⁷ J  

Intensity of light = 5 x 10⁻¹² J /m²

Total energy flowing through eye per second

= intensity x area of eye.

=   5 x 10⁻¹² x π x .2² x 10⁻⁴

= .628 x 10⁻¹⁶ J

so

n x  .036 x 10⁻¹⁷  =  .628 x 10⁻¹⁶

n = .628 x 10⁻¹⁶ /  .036 x 10⁻¹⁷

= 17.36 x 10

= 173.6

174.

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If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is ha
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false.

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Remember that the velocity is a vector, so it has a direction.

Then when she goes from the 1st end to the other, the velocity is positive

When she goes back, the velocity is negative

if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:

AV = (V + (-V))/2  = 0

While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.

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2 years ago
A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular vel
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Answer:

The net torque is 0.0372 N m.

Explanation:

A rotational body with constant angular acceleration satisfies the kinematic equation:

\omega^{2}=\omega_{0}^{2}+2\alpha\Delta\theta (1)

with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha

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\alpha=-\frac{-(15.1)^{2}}{2(114.3)}=1.00\frac{rad}{s^{2}}

The negative sign indicates the sphere is slowing down as we expected.

Now with the angular acceleration we can use Newton's second law:

\sum\overrightarrow{\tau}=I\overrightarrow{\alpha} (2)

with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:

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With M the mass of the sphere an R its radius, then:

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Then (2) is:

\sum\overrightarrow{\tau}=0.037(-1.00)=0.037 Nm

7 0
2 years ago
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