Answer:
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Explanation:
To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.
Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb
Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation 


=65.44 rad/sec


= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio 
also
transmission ratio ![= \frac{1}{[\frac{w}{w_n}]^{2} -1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%5B%5Cfrac%7Bw%7D%7Bw_n%7D%5D%5E%7B2%7D%20-1%7D)
![0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}](https://tex.z-dn.net/?f=0.7%20%3D%5Cfrac%7B1%7D%7B%5B%5Cfrac%7B65.44%7D%7Bw_n%7D%5D%5E2%20-1%7D)
SOLVING FOR Wn
Wn = 42 rad/sec

k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m
Answer:
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Answer:
Explanation:
Given
Initial Thickness=45 mm
Final thickness=20 mm
Roll diameter=600 mm
Radius(R)=300 mm
coefficient of friction between rolls and strip (
)=0.15
maximum draft

Minimum no of passes
(b)draft per each pass


Answer:
A. $28
Explanation:
We have the load to be 240V
6000W will be converted to kilowatts
6000/1000 = 6kW
Time heater runs for 36 hours
Cost of electricity = $0.13/Kwh
The question needs us to find cost of running this electric heater
Total kilowatts
= 6kw x 36 hours
= 216kwh
Then the total cost = 216 x $0.13
= 28.08
= $28