At what point will the currents that transfer heat throughout a gas cease to flow?
2 answers:
You might be interested in
Answer:
A. The work done by the intern is 792 J.
B. The velocity of the gurney when it has moved 12 m is 6.1 m/s.
C. The 12-m journey takes 3.8 s.
Explanation:
Hi there!
Please see the attached figure for a description of the situation.
<u>Part A: </u>
Work is done by a force when it is applied in the direction of the displacement or against it. In this case, the only force applied in the direction of displacement is the horizontal component of the force applied by the intern.
By trigonometry, the horizontal component of the force is calculated as follows:
cos θ = adjacent/hypotenuse
Looking at the figure, you can notice that the applied force, F, is the hypotenuse of a right triangle and the horizontal component, Fx, is the adjacent side:
cos θ = Fx / F
Fx = F · cos θ
Fx = 80 N · cos 35°
Fx = 66 N
Now we can calculate the work (W) done by this force:
W = Fx · x
Where x is the displacement:
W = 66 N · 12 m = 792 J
The work done by the intern is 792 J.
<u>Part B:</u>
Applying the work-energy theorem, the work done is equal to the change in kinetic energy:
W = final kinetic energy - initial kinetic energy
Since the gurney starts from rest, the initial kinetic energy is zero. Then:
W = final kinetic energy
The equation of kinetic energy (KE) is the following:
KE = 1/2 · m · v²
Where:
m = mass of the gurney.
v = velocity.
KE = 792 J
792 J = 1/2 · 42 kg · v²
v²= 2· 792 J / 42 kg
v = 6.1 m/s
The velocity of the gurney when it has moved 12 m is 6.1 m/s.
<u>Part C:</u>
First, let´s find the acceleration of the gurney:
Fx = m · a
Fx/m = a
66N / 42 kg = a
a = 1.6 m/s²
Now using the equation of velocity, let´s find the time at which the gurney has a velocity of 6.1 m/s:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = accleration.
t = time.
v = v0 + a · t
6.1 m/s = 0 + 1.6 m/s² · t
t = 6.1 m/s / 1.6 m/s²
t = 3.8 s
The 12-m journey takes 3.8 s.
Answer:
(C) 16
Explanation:
Given:
The amplitude of first wave (s₁) = 20 mm
The amplitude of second wave (s₂) = 5 mm
Intensity of first wave = Iₓ
Intensity of second wave =
The intensity associated with a wave depends on the amplitude of the wave.
The intensity (I) is directly proportional to the square of the amplitude (s) of the wave and is expressed as:
Now, the intensities of the two waves are given as:
Dividing both the intensities, we get:
Therefore, the option (C) is correct.