Answer:
Explanation:
Hi there,
To get started, please recall what heating curves are, latent heat, and enthalpy of phase changes (vaporization, fusion).
In a heating curve for ice water, temperature is the dependent variable (y-axis) and heat or time elapsed is independent variable (x-axis). There are three sloped portions. These sloped portions are showing where the <u>mixture is not changing phase, and is in a specific phase.</u> (solid first, then liquid then gas).
The reason is because the addition of more heat in time is causing the temperature to rise, so the mixture is getting "hotter." The temperature is still rising, yet the mixture is still in the current phase (depending on where you are referencing on the curve).
The two leveled portions are the process of melting (first leveled) and boiling (second leveled). The reason they are horizontal is because the addition of heat is not causing temperature to go up. The molecules are in the process of distributing heat in a different form by changing the mixtures entire structure (AKA phase), and so they are flat because the average kinetic energy from heat is being used change phase.
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Hope you do well on the test and hope this helps!
Continental air masses are characterized by dry air near the surface while maritime air masses are moist. Polar air masses are characterized by cold air near the surface while tropical air masses are warm or hot. ... Maritime polar (mP) air masses are cool, moist, and unstable.
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
Hi there!
a.
We know that:
Begin by determining the forces in the vertical direction:
W = weight of traffic light
T₁sinθ = vertical component of T₁
T₂sinθ = vertical component of T₂
b.
The ropes provide a horizontal force:
T₁cosθ = Horizontal component of T1
T₂cosθ = Horizontal component of T2
Thus:
0 = T₁cosθ - T₂cosθ
T₁cosθ = T₂cosθ
T₁ = T₂
c.
Since the angles for both ropes are the same, we can say that:
T₁ = T₂
Sum the forces:
ΣFy = T₁sinθ + T₁sinθ - W = 0
2T₁sinθ = W
d.
Now, we can begin by solving for the tensions:
2T₁sinθ = W
