In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of δ
1 answer:
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Answer:
<em>The time interval required to lift the spacecraft to this specified height is 123.94 seconds</em>
Explanation:
Height through which the spacecraft is to be lifted = 32.0 m
Mass of the spacecraft = 260.0 kg
Four crew member each pull with a power of 135 W
18.0% of the mechanical energy is lost to friction.
work done in this situation is proportional to the mechanical energy used to move the spacecraft up
work done = (weight of spacecraft) x (the height through which it is lifted)
but the weight of spacecraft = mg
where m is the mass,
and g is acceleration due to gravity 9.81 m/s
weight of spacecraft = 260 x 9.81 = 2550.6 N
work done on the space craft = weight x height
==> work = 2550.6 x 32 = 81619.2 J
this is equal to the mechanical energy delivered to the system
18.0% of this mechanical energy delivered to the pulley is lost to friction.
this means that
0.18 x 81619.2 = <em>14691.456 J </em> is lost to friction.
Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74<em> J</em>
Total power delivered by the crew to do this work = 135 x 4 = 540 W
But we know tat power is the rate at which work is done i.e
where p is the power
where w is the useful work done
t is the time taken to do this work
imputing values, we'll have
540 = 66927.74/t
t = 66927.74/540
time taken t = <em>123.94 seconds</em>
Answer:
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge
where Distance between the two charges
negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge
The magnitude of force by both the charge is same but at an angle of
thus combination of two forces at 2 and 3 will be
Now it will add with force due to 1 charge
Thus net force will be
