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Cerrena [4.2K]
3 years ago
11

A bogie is one more stroke over par true or false in footgolf

Physics
1 answer:
bezimeni [28]3 years ago
5 0

Answer:

True

Explanation:

I play

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To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 T
Ronch [10]

Correct question is;

To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright?

Answer:

We'll see more sharply in dim light

Explanation:

If we consider diffraction through a circular aperture, then angular resolution is given by;

θ = 1.22λ/D

where:

θ is the angular resolution (radians) λ is the wavelength of light

D is the diameter of the lens' aperture.

Thus,

at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm

θ = (1.22 * 550)/(2 x 10^(6))

θ = 335.5 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]

= 69.18 arc seconds

at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm

θ = (1.22 * 550)/(8 x 10^(6))

θ = 83.875 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]

= 17.3 arc seconds

From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.

4 0
3 years ago
In an RC series circuit, ε = 12.0 V, R = 1.25 MΩ, and C = 1.42 µF. (a) Calculate the time constant. (b) Find the maximum charge
faust18 [17]

Answer:

a, 1.775s

b, 17.04μC

c, 1.28s

Explanation:

Given

R = 1.25MΩ

C = 1.42µF

ε = 12.0 V

q = 8.78 µC

Time constant, τ = RC

τ = (1.25*10^6) * ( 1.42*10^-6)

τ = 1.775s

q• = εC

q• = 12 * 1.42*10^-6

q• = 17.04*10^-6C

q• = 17.04μC

Time t =

q = q• [1 - e^(t/τ)]

t = τIn[q•/(q•-q)]

t = 1.775In[17.04μC/(17.04μC-8.78μC)]

t = 1.775In(2.06)

t = 1.775*0.723

t = 1.28s

8 0
3 years ago
Could you help me with a science question really quick?
Rama09 [41]
Yea I would love to help u
4 0
2 years ago
A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the sk
malfutka [58]

Answer:

The mass of the cargo is M  =  188.43 \ kg

Explanation:

From the question we are told that

    The radius of the spherical balloon is  r =  7.40 \ m

     The mass of the balloon is  m = 990\ kg  

The volume of the spherical balloon is mathematically represented as

     V  =  \frac{4}{3} * \pi r^3

substituting values

      V  =  \frac{4}{3} * 3.142 *(7.40)^3

      V  =  1697.6 \ m^3

The total mass  the balloon can lift is mathematically represented as

     m =  V (\rho_h - \rho_a)

where \rho_h is the density of helium with a  value of

       \rho_h  =  0.179 \ kg /m^3

and  \rho_a is the density of air with a value of

        \rho_ a  = 1.29 \ kg / m^3

substituting values

          m =  1697.6 ( 1.29  - 0.179)

         m =  1886.0  \ kg

Now the mass of the cargo is mathematically evaluated as

        M  =  1886.0 - 1697.6

        M  =  188.43 \ kg

       

5 0
3 years ago
An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free
Blababa [14]

Answer:

the free-fall acceleration on the moon is 1.68 m/s^2

Explanation:

recall the formula for the gravitational potential energy (under acceleration of gravity "g"):

PE = m * g * h

replacing with our values for the problem:

46 J = 91 * g * 0.3

solve for the "g" on the Moon:

g = 46 / (91 * 0.3)

g = 1.68  m/s^2

3 0
2 years ago
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