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Arturiano [62]
3 years ago
8

The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e

yes of air-dwelling mammals. In mammalian eyes, the shape (and therefore the focal length) of the lens changes to enable the eye to focus at different distances. In amphibian eyes, the shape of the lens doesn't change. Amphibians focus on objects at different distances by using specialized muscles to move the lens closer to or farther from the retina, like the focusing mechanism of a camera. In air, most frogs are nearsighted; correcting the distance vision of a typical frog in air would require contact lenses with a power of about −6.0D .A frog can see an insect clearly at a distance of 10cm . At that point the effective distance from the lens to the retina is 8mm .If the insect moves 5cm farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus?0.02cm, toward the retina.0.02cm, away from the retina.0.06cm, toward the retina.0.06cm, away from the retina.
Physics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

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What are the two main types of energy?
djyliett [7]

Answer:

Fourth option

Explanation:

They're many different types of energy, from chemical and mechanical to heat and solar energy. But the two most basic types of energy are "kinetic and potential energy" or the fourth option. Kinetic energy is the energy an object has when it is in motion, while potential energy is the energy an object has when it's as rest. These two specific types of energies are the most basic and you can even convert them into many different types of energies, like heat or electrical energy.

Hope this helps.

7 0
2 years ago
Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/sthrougha horizontal tube with a 30mmdiameter. what
marusya05 [52]

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>
  • We have the Poiseuille's formula,

                     Q=\frac{\pi r^4P}{8\beta l}

  • where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.
  • By substituting values and rearranging we will get the pressure drop as,

                  P=3.284Pascals

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.

Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

#SPJ4

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1 year ago
Why doesn’t she love me
o-na [289]
I bet she does just give her tule work on yourself
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3 years ago
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marta [7]

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4 0
3 years ago
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
4 0
4 years ago
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