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Flauer [41]
3 years ago
13

A shallow hole is dug and lined with concrete. A number of large black plastic bags, each filled with several centimeters of wat

er, are placed in the hole and then covered with fiberglass insulation panels. The panels let sunlight in but keep most of the heat stored in the water during the daytime from being lost to the atmosphere. This heat can then be used to heat water and space. It cannot be used to generate electricity. This technology is called:
Physics
1 answer:
Liula [17]3 years ago
8 0

Answer:

Freshwater solar ponds or freshwater floating collector type solar pond.

Explanation:

A shallow hole is dug and lined with concrete. A number of large black plastic bags, each filled with several centimeters of water, are placed in the hole and then covered with fiberglass insulation panels. The panels let sunlight in but keep most of the heat stored in the water during the daytime from being lost to the atmosphere. This heat can then be used to heat water and space. It cannot be used to generate electricity. This technology is called freshwater solar ponds or freshwater floating collector type solar pond.

  • This type of solar pond technology is introduced to provide a cheap, renewable heat source which is used mainly for space - heating purposes. The freshwater solar pond is covered with floating collectors/ fiberglass insulation panels which are made of thin plastic sheet, insulation board and it also requires a little constructional work.
  • In summary, the collector also provide thermal and evaporation insulation so that fresh water may be used when the pond gives reduction of mean collector and pond temperature and ensures high efficiency through forced flow in collector layer of pond.

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1. Bone has a Young’s modulus of about
Blababa [14]

#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

\Delta L = 5.04 mm

#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

\rho = 1185.3 kg/m^3

#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

46.6 = 43.89 - 922x + 1000x

x = 3.48 cm

so it is 3.48 cm below the interface

5 0
3 years ago
A passenger jet travels from Los angles to Bombay, India, in 22 h. The return flight takes 17 h. The difference in flight times
Hoochie [10]

Answer:

70.5 mph

Explanation:

A passenger jet travels from Los Angeles to Bombay, India, in 22h.

The return flight takes 17 h.

The difference in flight times is caused by winds over the Pacific Ocean that

blow primarily from west to east.

If the jet's average speed in still air is 550 mi/h what is the average speed

of the wind during the round trip flight? Round to the nearest mile per hour.

Is your answer reasonable?

:

Let w = speed of the wind

:

Write a distance equation (dist is the same both ways

17(550+w) = 22(550-w)

9350 + 17w = 12100 - 22w

17w + 22w = 12100 - 9350

39w = 2750

W = 2750/39

w = 70.5 mph seems very reasonable

:

Confirming if the solution by finding the distances using these value

17(550+70.5) = 10549 mi

22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph

6 0
3 years ago
A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro
NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width  = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

5 0
3 years ago
Convert 1.5 radian into milliradians.
Alex

Answer:

1500 milliradians

Explanation:

Data provided in the question:

1.5 radians

Now,

1 radians consists of  1000 milliradians

1 milli = 1000

thus for the 1.5 radians, we have

1.5 radians = 1.5 multiplied by 1000 milliradians

or

1.5 radians = 1500 milliradians

Hence, after the conversion

1.5 radians equals to the value 1500 milliradians

4 0
3 years ago
In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from
Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s     time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s     time =t      

we know,

distance = speed x time

distance of BC

using equation of motion

v = u + a t

27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

s = ut + \dfrac{1}{2}at^2

s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2

s = 68.94 m

3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}

3000 = 27.5 t + 68.94 + 13.33 t

40.83 t = 2931.06

t = 71.79 s

distance travel in AB

distance = s x t

d = 27.22 x 71.79

d = 1954 m

d = 1.954 Km

distance between A and B is equal to 1.954 Km.

4 0
3 years ago
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