The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;
V(in³) = (¹/₆₁₀₂₄)a
We are given that;
Volume of microbial culture is observed to increase according to the formula;
V = e^(t)
where;
t is in seconds
V is in m³
We want to now express V in in³ and t in minutes.
Now, from conversions;
1 m³ = 61024 in³
Also; 1 second = 1/60 minutes
according to formula for exponential decay, we know that;
V = ae^(bt)
Thus, we have;
61024V = ae^(¹/₆₀b(t(h))
V(in³) = (¹/₆₁₀₂₄)a
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Answer:
Explanation:
Given
charge is placed at 
another charge of
is at 
We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.
so net electric field is zero somewhere beyond negatively charged particle
Electric Field due to
at some distance r from it

Now Electric Field due to
is

Now 



thus 
Thus Electric field is zero at some distance r=1.43 cm right of
Answer:
Change in entropy S = 0.061
Second law of thermodynamics is satisfied since there is an increase in entropy
Explanation:
Heat Q = 300 kW
T2 = 24°C = 297 K
T1 = 7°C = 280 K
Change in entropy =
S = Q(1/T1 - 1/T2)
= 300(1/280 - 1/297) = 0.061
There is a positive increase in entropy so the second law is satisfied.
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.