Answer: It will be take 2.6 hours
Explanation: Please see the attachments below
Answer:
0.60 m/s
Explanation:
The average velocity from t = a to t = b is:
v_avg = (x(b) − x(a)) / (b − a)
Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed is 0.60 m/s.
Answer: 9130 joules
Explanation:
Workdone by wheelbarrow = ?
Time = 11 seconds
Power = 830 watts
Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.
i.e Power = (workdone/time)
830 watts = Workdone / 11 seconds
Workdone = 830 watts x 11 seconds
Workdone = 9130 joules
Thus, 9130 joules of work is required to get the wheelbarrow across the yard.
Answer:
The work done on the gas is equal to the area under the curve pv diagram w = area of triangle = 1/2 (base)(height) = 1/2 (BC)(Ac) = 1/2 (3v - v)(3p - p) = 1/2 (9 vp - 3 vp - 3vp + vp) = 4 vp/2 W = 2 vp
Check attachment for the diagrammatic representation