Answer:
The expected dynamic error is 0.019
The phase shift is -23.10°C
Explanation:
The explanation is shown on the first uploaded image
A cable that weighs 4 lb/ft is used to lift 1000 lb of coal up a mine shaft 700 ft deep. Find the work done.
1 answer:
Answer:
<h2>980000ft-lbs</h2>Explanation:
Step one:
given data
mass of cable= 4lb/ft
mass of coal= 1000lb
dept of mine= 700ft
Step two:
Required
the work-done to lift the coal and the rope combined
Work-done to lift coal
Wc=1000*700= 700,000 lb-ft
Work-done to lift rope
substitute y=700 we have, since y=0 will result to 0
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Answer:
For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.
Explanation:
The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.
From the slope of graph it is clear that acceleration at t = 1 sec is given as:
Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2
Now, there are two cases:
1- Elevator moving up
2- Elevator moving down
For upward motion:
Apparent Weight = m(g + a)
Apparent Weight = (75 kg)(9.8 + 4)m/s^2
<u>Apparent Weight = 1035 N</u>
For downward motion:
Apparent Weight = m(g - a)
Apparent Weight = (75 kg)(9.8 - 4)m/s^2
<u>Apparent Weight = 435 N</u>
