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Ganezh [65]
4 years ago
15

The main difference difference ponds and lakes is that ponds

Chemistry
2 answers:
STALIN [3.7K]4 years ago
8 0

The difference is actually a result of the the depth. Ponds, according to limnology (the study of water bodies) are shallow enough where plants could conceivably grow across the entire surface. ... As a result, there are some very small bodies of water, less than an acre that are deep enough to be called lakes.

Flauer [41]4 years ago
5 0

The difference is actually a result of the the depth. Ponds, according to limnology (the study of water bodies) are shallow enough where plants could conceivably grow across the entire surface. ... As a result, there are some very small bodies of water, less than an acre that are deep enough to be called lakes. Both are small bodies of water, either natural or man-made, that are completely surrounded by land. The primary difference between the two is their size. Simply put, lakes are larger and ponds are smaller. However, there is no standardization of lake sizes.


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Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow
Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

   Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

         \frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)

      \rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h

   \frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}

          [/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = \frac{10}{\rho A_{c}}

                       A_{c} = 0.01 m^{2}

              \frac{dh}{dt} + h = 1

                  \frac{dh}{dt} = 1 - h

               \frac{dh}{1 - h} = dt  

                \frac{ln(1 - h)}{-1} = t + C      

Given at t = 0 and V = 0  

                         A \times h = 0  

 or,                     h = 0

                 -ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

                                C = 0  

                So,   -ln(1 - h) = t

or,                      t = ln (\frac{1}{1 - h})  ........... (1)

(a)    Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

                    t = ln (\frac{1}{1 - h})  

                     t = ln (\frac{1}{1 - 0.6})  

                        = ln (\frac{1}{0.4})

                        = 0.916 seconds

(b)   As maximum height of water level in the tank is achieved at steady state that is, t = \infty.  

                    1 - h = exp (-t)

                    1 - h = 0  

                         h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

                 

8 0
3 years ago
8 points
butalik [34]

Answer:

1.3 meters

Explanation: use newton third law equation.

5 0
3 years ago
5. What type of clouds can be found in the mesosphere?
Masja [62]

Some material from meteors lingers in the mesosphere, causing this layer to have a relatively high concentration of iron and other metal atoms. Very strange, high altitude clouds called "noctilucent clouds" or "polar mesospheric clouds" sometime form in the mesosphere near the poles.

I really hope this helps! I wish you the best of luck!

8 0
3 years ago
5.4x10^3x1.2 x 10^7 in scientific notation
BartSMP [9]

6.4*10^10 is the answer in scientific notation


5 0
3 years ago
Cu20(s) + C(s) - 2Cu(s) + CO(g)
Romashka-Z-Leto [24]

Answer:

That means Cu2O is limiting reagent and C is excess reagent

Explanation:

Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.

To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.

<em>Moles Cu2O -Molar mass: 143.09 g/mol-</em>

114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O

<em>Moles C -Molar mass: 12.01g/mol-</em>

11.1g C * (1mol / 12.01g) = 0.924 moles C

<h3>That means Cu2O is limiting reagent and C is excess reagent</h3>

<em> </em>

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6 0
3 years ago
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