John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv
2 answers:
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Answer:
by ideal gas pressure = 15529.475 kPa
by compressibility chart pressure = 12576 kPa
by steam tables Pressure = 12517 kPa
Explanation:
given data
temperature T = 400°C = 673 K
volume v = 0.02 m³/kg
to find out
pressure by ideal gas, compressibility chart and steam tables
solution
we know here by table
gas constant R is 0.4615 kJ/ kg-K
and critical temp Tc = 647.1 K
and critical pressure Pc = 22064 kPa
so by ideal gas pressure is
pressure = R×T / v
pressure = 0.4615 × 673 / 0.02
pressure = 15529.475 kPa
and
by compressibility chart
temperature reduce is = T/ Tc
temperature reduce Tr = 673 / 647.1
Tr = 1.040 K
so pseudo reduce volume is here
reduce volume Vr = v / ( RTc/Pc)
reduce volume Vr =
0.02 / ( 461.5(647.1) / 22064×10³)
reduce volume = 1.48
and we know by compressibility chart
reduce pressure Pr is 0.57
so
pressure = Pr × Pc
pressure = 0.57 × 22064 × 10³
pressure = 12576 kPa
and
from steam table
pressure is 12.5 MPa at 673 K and 0.020030 m³/kg
pressure is 15 MPa at 673 K and 0.015671 m³/kg
so
pressure P is
=
so
Pressure = 12517 kPa
Answer:
a = 3.27 m/s²
F = 32.7 N
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling straight down.
Normal force N pushing perpendicular to the slope.
Friction force F pushing parallel up the slope.
Sum of forces in the parallel direction:
∑F = ma
mg sin θ − F = ma
Sum of torques about the cylinder's axis:
∑τ = Iα
Fr = ½ mr²α
F = ½ mrα
Since the cylinder rolls without slipping, a = αr. Substituting:
F = ½ ma
Two equations, two unknowns (a and F). Substituting the second equation into the first:
mg sin θ − ½ ma = ma
Multiply both sides by 2/m:
2g sin θ − a = 2a
Solve for a:
2g sin θ = 3a
a = ⅔ g sin θ
a = ⅔ (9.8 m/s²) (sin 30°)
a = 3.27 m/s²
Solving for F:
F = ½ ma
F = ½ (20 kg) (3.27 m/s²)
F = 32.7 N
Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
To get the velocity, we integrate over time:
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
To get the position equation we integrate again over time:
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
To get the position equation, we can integrate:
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
We can solve this for t.
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
We only care for the positive solution, which is:
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s