An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and
1 answer:
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The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle 
between the direction of the force and the direction of the displacement:
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so
Therefore, the work done is
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so
Therefore, the work done is
Remain the same
Explanation:
If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.
Work done is the force applied to move a body through a distance.
Work done = F x d
where F is the applied force
d is the distance moved
Now;
if:
f = 2f
d = d
Input the parameter:
Work done = fxd = 2f x d = fd
The work done will still remain the same
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Answer:
(a)
(b)
Explanation:
Hello.
(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:
Which is negative as it expands against the given pressure.
(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:
Whereas the moles are computed from the given mass of argon:
Thus, the work is:
Regards.