A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it
1 answer:
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Answer:
new boiler total cost = $229706.825
new boiler total cost = $127512
Explanation:
given data
power p1 = 80 kW
cost C = $160000
cost index CI 1 = 187
cost index CI 2= 194
cost capacity factor f = 0.6
power p2 = 120 kW
current cost = $18000
to find out
total cost and cost of 40 kW
solution
we consider here CN cost for new boiler and CO cost for old boiler
and x is capacity of new boiler
first we find old boiler current cost that is
current cost CO = C × .............1
put here value
current cost = 160000 ×
new current cost = $165989.304
and
use here power sizing technique for 124 kW
CN/CO = ...............2
put here value and find CN
CN/CO =
CN / 165989.304 =
CN = 211706.825
so new cost = $211706.825
so
total cost for new boiler is
total cost = new cost + current cost
total cost = 211706.825 + 18000
new boiler total cost = $229706.825
and
for 40 kW new cost will be
use equation 2
CN/CO =
CN / 165989.304 =
CN = 109512
so new cost is $109512
so
total cost for new boiler is
total cost = new cost + current cost
total cost = 109512 + 18000
new boiler total cost = $127512
Answer:
ΔKE=-347.278 kJ
ΔPE= 441.45 kJ
Explanation:
given:
mass m=900 kg
the gravitational acceleration g=9.81 m/s^2
the initial velocity =100 km/h-->100*10^3/3600=27.78 m/s
height above the highway h=50 m
h1=0m
the final velocity =0 m/s
<u>To find:</u>
the change in kinetic energy ΔKE
the change in potential energy ΔPE
<u>assumption:</u>
We take the highway as a datum
<u>solution:</u>
ΔKE=5*m*(^2-
^2)
=-347.278 kJ
ΔPE=m*g*(h-h1)
= 441.45 kJ
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
