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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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En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8
Ivanshal [37]

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

A=10cm\\T=2s

the equation of the motion is

x=Acos(\omega t)\\

we can calculate w by using

\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi

Hence, we have that

x=10cm*cos(\pi t)\\

the speed will be

v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}

hope this helps!

6 0
3 years ago
15. Find the speed of a disc of radius 0.5 meters and mass 2-kg at the base of the incline. The disc starts at rest and rolls do
mote1985 [20]

Explanation:

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3 years ago
You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what
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3 years ago
A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at
riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0
3 years ago
A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the
sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0
3 years ago
Read 2 more answers
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