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Alex
3 years ago
6

How is a warm front formation diffrent from occlued front formation​

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:  Two cold air masses surround a warm air mass during a warm front, but a cold air mass moves against a warm air mass during an occluded front.

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ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved.
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Answer:

Therefore, the number of grams of salt in the tank at time t is A(t) = 150-140 e^{-\frac{t}{30} }

Explanation:

Given:

Tank A contain V_{1} = 150 lit

Rate \alpha = 5 \frac{L}{min}

Dissolved salt A = 10 gm

Salt pumped in one minute is 4 \frac{L}{min}

Salt pumped out is \frac{5L}{150L} = \frac{1}{30} of initial amount added salt.

To find A(t)

  \frac{dA}{dt}  = Rate _{in} - Rate _{out}

  A' = 5 - \frac{A}{30}

  A' + \frac{A}{30} = 5

Solving above equation,

  I .F   = e^{\int\limits {p} \, dt }

   y = e^{\int\limits {\frac{1}{30} } \, dt }

   y = e^{\frac{t}{30} }

(Ae^{\frac{t}{30} }  )' = 5 e^{\frac{t}{30} } + c

Integrating on both side,

Ae^{\frac{t}{30} } =  5 \times 30 e^{\frac{t}{30} } +c

Add e^{-\frac{t}{30} } on above equation,

 A = 150 + ce^{-\frac{t}{30} }

Here given in question,A(t=0) = 10

 10 =150 +c

   c = -140

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

 A(t) = 150-140 e^{-\frac{t}{30} }

Therefore, the number of grams of salt in the tank at time t is A(t) = 150-140 e^{-\frac{t}{30} }

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The diving velocity of a falcon can reach 322 km/hr. If the falcon maintains this velocity for 7.2 seconds (0.002 hours), what w
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Hope this helped!
7 0
4 years ago
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