Hi there!
(a)
Recall that:
W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.
To the nearest multiple of ten:
(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:
(c)
Similarly, the normal force is perpendicular to the displacement, so:
(d)
Recall that the force of kinetic friction is given by:
Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
In multiples of ten:
(e)
Simply add up the above values of work to find the net work.
Nearest multiple of ten:
(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
Nearest multiple of ten:
Answer:
"Offgassing"
Explanation:
According to my research on Kinesiology, I can say that based on the information provided within the question the process being described is known as "Offgassing". In other words this process is defined as when something gives off or releases a chemical, especially a harmful one, in the form of a gas into the air..
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
The answer is D I took the test
Answer:
19.08 m/s
Explanation:
f = actual frequency emitted by the parked car's horn = 440 Hz
V = speed of sound = 342 m/s
f' = frequency of the horn observed by you = 466 Hz
v = speed of your car moving towards the parked car = ?
frequency of the horn observed by you is given as
v = 19.08 m/s
Answer:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings
