becky visited the great pyramid of Giza she learned that the pyramid was originally 481 feet tall and it’s square base has size
1 answer:
You might be interested in
Answer:
a) (dy/dt) = 2 - [3y/(100 + 2t)]
b) The solved differential equation gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L
Step-by-step explanation:
First of, we take the overall balance for the system,
Let V = volume of solution in the tank at any time
The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)
The rate of change of the volume of solution = dV/dt
Rate of flow into the tank = Fᵢ = 5 L/min
Rate of flow out of the tank = F = 3 L/min
(dV/dt) = Fᵢ - F
(dV/dt) = (Fᵢ - F)
dV = (Fᵢ - F) dt
∫ dV = ∫ (Fᵢ - F) dt
Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t
V - 100 = (Fᵢ - F)t
V = 100 + (5 - 3)t
V = 100 + (2) t
V = (100 + 2t) L
Component balance for the amount of salt in the tank.
Let the initial amount of salt in the tank be y₀ = 0 kg
Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min
Amount of salt in the tank, at any time = y kg
Concentration of salt in the tank at any time = (y/V) kg/L
Recall that V is the volume of water in the tank. V = 100 + 2t
Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min
Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)
(dy/dt) = 2 - (3y/V)
(dy/dt) = 2 - [3y/(100 + 2t)]
To solve this differential equation, it is done in the attached image to this question.
The solution of the differential equation is
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration after 20 minutes.
After 20 minutes, volume of water in tank will be
V(t) = 100 + 2t
V(20) = 100 + 2(20) = 140 L
Amount of salt in the tank after 20 minutes gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵
y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵
y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵
y(20) = 56 - 24.15 = 31.85 kg
Amount of salt in the tank after 20 minutes = 31.85 kg
Volume of water in the tank after 20 minutes = 140 L
Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L
Hope this Helps!!!
Answer:
Step-by-step explanation:
we know that
To Round a number
a) Decide which is the last digit to keep
b) Leave it the same if the next digit is less than (this is called rounding down)
c) But increase it by if the next digit is
or more (this is called rounding up)
In this problem we have
We want to keep the digit
The next digit is which is greater or equal than
, so increase the
by 1 to
therefore
the answer is
The answer is going to be A.) -2
