Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
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<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
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<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
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<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
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.
Answer:
Reduce manufacturing costs.
Explanation:
Hope This Helps
Have A Great Day
Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar = 
ar = 
ar = 22.74 m/s²
so magnitude of total acceleration is
A = 
A = 
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²
