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Arturiano [62]
2 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]2 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.

You might be interested in
The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
2 years ago
An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo
lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

\frac{d'_1}{d'_2}=\frac{f_1}{f_2}

Magnification of first lens

M_2=-\frac{d'_1}{d_1}

                   and

M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

                   and

\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

which means,

\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

and

\frac{d'_1}{d_2'}=\frac{f_1}{f_2}

So, we get

\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm

∴ Focal length should this camera's lens is 24.71 mm

6 0
2 years ago
100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak
Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

Q_m=AC_o\sqrt{2\rho g_cP_g}

Here, Q_m is the mass flow rate through the leak of the pipeline. A is the area of the hole, C_o is the discharge rate, \rho is the fluid density, g_c is the gravitational constant and P_g is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft

Calculate the area (A) of the hole. The area of the hole is.

A=\frac{\pi d^2}{4}

Substitute 3.14 for \pi and 8.33\times 10^{-3}ft for d and calculate A.

A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is 62.4\frac{Ib_m}{ft^3}. Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

P_g=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate P_g.

Upstream x distance from upstream pressure end

P_g=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig

Convert the pressure from psig to Ib_f/ft^2

P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute 5.45\times 10^{-5}ft^2 for A. 0.61 for C_o, 54.9\frac{Ib_m}{ft^3} for \rho, 32.17\frac{ft.Ib_m}{Ib_f.s^2} for g_c, and 6,379.2\frac{Ib_f}{ft^2} for P_g and calculate Q_m

Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}

The mass flow rate of benzene through the leak in the pipeline is 0.158\frac{Ib_m}{s}

8 0
2 years ago
James gently releases a ball at the top of a slope, but does not push the ball. The ball rolls down the slope. Which force cause
Rzqust [24]

D. Gravitational force.

5 0
2 years ago
Read 2 more answers
A piano string sounds a middle A by vibrating primarily at 220 Hz.a)Calculate its period.b)Calculate its angular frequency.c)Cal
chubhunter [2.5K]

a) 4.5 ms

The period of a wave is given by:

T=\frac{1}{f}

where f is the frequency.

For the note in this problem, f = 220 Hz, so the period of the wave is

T=\frac{1}{f}=\frac{1}{220 Hz}=4.5\cdot 10^{-3} s = 4.5 ms

b) 1381.6 rad/s

The angular frequency is given by:

\omega=2 \pi f

where f is the frequency.

In this problem, f = 220 Hz, so the angular frequency is

\omega=2 \pi (220 Hz)=1381.6 rad/s

c) 1.1 ms

The frequency of the "high A" is four times the frequency of the piano string, so

f=4 \cdot 220 Hz=880 Hz

And so, its period is

T=\frac{1}{f}=\frac{1}{880 Hz}=1.1\cdot 10^{-3} s=1.1 ms

d) 5526.4 rad/s

The angular frequency is given by:

\omega=2 \pi f

where f is the frequency.

For this note, f = 880 Hz, so the angular frequency is

\omega=2 \pi (880 Hz)=5526.4 rad/s

5 0
2 years ago
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