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lisov135 [29]
4 years ago
13

What is a cone of depression

Physics
1 answer:
Thepotemich [5.8K]4 years ago
7 0

A Cone of Depression occurs when a well pumps water from a certain area and causes a depression in a cone shape in the water table.


Hope this helps! :)


if you get it correct Brainliest is greatly appreciated

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What is the equivalent of 0° C in Kelvin?<br> Help will give brainiest
givi [52]

Answer:

273.15 dat is the answer im pretty sure

8 0
3 years ago
Q 24, 25, 26 i dont get them and need answers for it
Alexandra [31]

Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

DU = Q-W\\

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]  

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

Q = W\\

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

E_{p}=Q\\\\E_{p}=m*g*h

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]

The heat developed can be calculated by means of the following equation.

Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

7 0
4 years ago
Help me!!!!!!!!!!!!!!!!!!!!!!!
Gemiola [76]

Answer:a computer , machine forcery,0,push

Explanation:

8 0
3 years ago
Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0
3 years ago
Give 5 examples of physical properties that could be used to describe a solid
ycow [4]
1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor
8 0
3 years ago
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