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m_a_m_a [10]
3 years ago
15

A cylindrical cistern, constructed below ground level, is 2.9 m in diameter and 2.0 m deep and is filled to the brim with a liqu

id whose index of refraction is 1.4. A small object rests on the bottom of the cistern at its center. How far from the edge of the cistern can a girl whose eyes are 1.2 m from the ground stand and still see the object
Physics
1 answer:
Tomtit [17]3 years ago
4 0

Answer:

15.1 m

Explanation:

We first calculate the apparent depth from

refractive index, n = real depth/apparent depth

apparent depth, a  = real depth/refractive index

real depth = 2.0 m, refractive index = 1.4

apparent depth, a = 2.0/1.4 = 1.43 m

Since the cylindrical cistern has a diameter of 2.9 m, its radius is 2.9/2 = 1.45 m

The angle of refraction, r is thus gotten from the ratio

tan r = radius/apparent depth = 1.45/1.43 = 1.014

r = tan⁻¹(1.014) = 45.4°

The angle of incidence, i is gotten from n = sin i/sin r

sin i = nsin r = 1.4sin45.4° = 1.4 × 0.7120 = 0.9968

i = sin⁻¹(0.9968) = 85.44°

Since the girl's eyes are 1.2 m from the ground, the distance ,h from the edge of the cistern she must stand is gotten from

tan i = h/1.2

h = 1.2tan i = 1.2tan85.44° = 1.2 × 12.54 = 15.05 m

h = 15.05 m ≅ 15.1 m

So, she must stand 15.1 m away from the edge of the cistern to still see the object.

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Answer:

13.37 rev/min

Explanation:

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r = 9 m

Centripetal acceleration (a_c) is given by:

a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s

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4 0
2 years ago
If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil
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Answer:

Explanation:

Let the volume of air be V. at atmospheric pressure, that is 10⁵  Pa

At 20 m below surface pressure will be

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10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa

At this pressure volume V becomes V/ 2.96  

This volume will last 1/2.96 times  time that is 60/2.96 = 20.27 minutes.

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3 years ago
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of
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Answer:

a) P=2450\ Pa

b) \delta h=23.162\ cm

Explanation:

Given:

height of water in one arm of the u-tube, h_w=25\ cm=0.25\ m

a)

Gauge pressure at the water-mercury interface,:

P=\rho_w.g.h_w

we've the density of the water =1000\ kg.m^{-3}

P=1000\times 9.8\times 0.25

P=2450\ Pa

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

\rho_w.g.h_w=\rho_m.g.h_m

1000\times 9.8\times 0.25=13600\times 9.8\times h_m

h_m=0.01838\ m=1.838\ cm

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\delta h=h_w-h_m

\delta h=25-1.838

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7 0
2 years ago
A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before sto
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Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

4 0
2 years ago
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