No, this is false.
In principle, every well can run dry, and this includes a well that has reached the water table. If the water is taken from the well faster than it's replenishing itself, the well will run dry- and in this situation, typically people need to look for a new place to have a well.
Answer:
Find the attachments for complete solution
Assuming that the gravitational field strength is 10 N/kg, the sum would look like this:
PEg=mgh
=60×10×10
=6000J
However, you will need to check that you are not meant to be using 9.8 N/kg as the gravitational field strength.
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
A. How much work is being done to hold the beam in place?
Work is the product of Force and Displacement. Since there
is no Displacement involved in just holding the beam in place, hence the work
is zero.
B. How much work was done to lift the beam?
In this case, force is simply equal to weight or mass
times gravity. Hence the work is:
Work = weight * displacement
Work = 500 lbf * 100 ft
Work = 50,000 lbf * ft
C. How much work would it take if the steel beam were
raised from 100 ft to 200ft?
The displacement is still 100 ft since 200 – 100 = 100 ft,
hence the work done is still similar in B which is:
<span>Work = 50,000 lbf * ft</span>
