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9966 [12]
2 years ago
6

Explain how a fish is able to enjoy 180° field of view when in a pond.​

Physics
1 answer:
marysya [2.9K]2 years ago
6 0

Answer:

Time

An explanation of basic pond measurements that are vital to basic pond management practices.

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Pond Measurements: Area, Volume and Residence Time - Articles UPDATED: OCTOBER 29, 2015

Pond Measurements: Area, Volume and Residence Time

The importance of getting an accurate estimation of your pond surface area cannot be overestimated. The majority of pond owners visually estimate their pond area, which usually results in an overestimate of the true pond surface area. Pond area and water volume should be calculated based on some simple measurements. The effort necessary to estimate pond surface area is directly related to your pond's shape and uniformity. The simplest method--using basic equations for common shapes--can be applied if your pond closely resembles a circle, square, rectangle, or trapezoid in shape.

Pond Shapes

Circular

pond shape can be estimated by measuring the distance around the pond shoreline in feet. Square the shoreline distance and divide by 547,390 to get the pond area in acres. For example, a pond that is 450 feet around the shoreline would have an area = (450 feet)2 / 547,390 or 0.37 acres.

Explanation:

i looked this up

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The number of players on a regulation Basketball Team?
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Read 2 more answers
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0
3 years ago
Explain the force of air resistance depends on an objects speed
Monica [59]

Answer:

more massive objects fall faster than less massive objects because they are acted upon by a larger force of gravity

Explanation:

3 0
3 years ago
Una mujer de masa m está parada en el borde de una mesa giratoria horizontal de momento de inercia I y radio R. La mesa al princ
Dovator [93]

r

- \frac{mR^2 }{I  } \ vAnswer:

a)      w = - \frac{m r }{I} v  ,  b)   W = - ½ m_woman R² (1 + m_woman R / I²) v²

Explanation:

a) To solve this exercise, let's use the conservation of angular momentum.

We define a system formed by the table and the woman, therefore the torques are internal and the moment is conserved

initial instant. Before starting to move the woman

         L₀ = 0

final instant. After starting to move

         L_f = I w + m v r

the moment is preserved

        L₀ = L_f

         0 = Iw + m v r

         w = - \frac{m r }{I} v                    (1)

the direction of the angular velocity is opposite to the direction of the linear velocity, that is, counterclockwise

b) for this part we use the relationship between work and kinetic energy

        W = ΔK

in this case the initial speed is zero and the final speed of the table, using the relationship between linear and angular variables

         v = w r

we substitute

          W = 0 - ½ I_total w²

          I_total = I + m_{woman} R²

          W = - ½ (I + m_woman R²)  ( \frac{m_{woman} R}{I} \ v) ²

          W = - ½ (m_woman² R² + m_woman³ R³ / I²) v²

          W = - ½ m_woman R² (1 + m_woman R / I²) v²

3 0
3 years ago
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