How can you do this 5.2.4: Rating?
1 answer:
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Answer:
work done by electric field is 0.06 J
Explanation:
Given data:
Two point charge is
0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)
-1 charge positioned is (0 cm , -1 cm, 0.00 cm)
E = 3.0\times 10^6 N/C
From above information, the distance between given two charges d = 2 cm
then d = 0.02m
work needed is W = q E d
W = 0.06 J
Therefore work done by electric field is 0.06 J
Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
<u><em>Power: </em></u>Rate of change of work with respect to time is called power.
We know that P=
rad/sec
So that P=
So 3281.08=
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb