If "0.3 minute" is correct, then it's 9,543,272 Joules.
If it's supposed to say "0.3 SECOND", then the KE is 2,651 Joules.
As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =<span>π</span><span>r*r</span>
d=120
<span>r=<span>120/2</span></span><span>r=60</span><span><span>60*60</span>=3600</span><span>3600*π=11309.734</span>
<span>11309.734 square miles</span>
Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.
Answer:
The effect of lowering the condenser pressure on different parameters is explained below.
Explanation:
The simple ideal Rankine cycle is shown in figure.
Effect of lowering the condenser pressure on
(a). Pump work input :- By lowering the condenser pressure the pump work increased.
(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.
(c). Heat supplied :- Heat supplied increases.
(d). Heat rejected :- The heat rejected may increased or decreased.
(e). Efficiency :- Cycle efficiency is increased.
(f). Moisture content at turbine exit :- Moisture content increases.
