Answer:
a = αR
Then We apply the Newton's second law of motion:
∑ F = ma
F - mg sinθ - f = ma.
F - mg sinθ - μmg cosθ = ma
Using given data we have:
F - 3217 = 470a ........................... (1)
Apply the Newton's law for rotation:
∑τ = I α
FR - (mg sinθ)R =
then: F - mg sinθ = 3ma / 2.
F - 2774 = 705a .......................... (2)
solving 1 and 2 we get:
F = 4103 N
a = 1.885 m/s²
b) In this case the time is given by:
t = it comes from: d =
starting from rest vo = 0
t = = 3.09 s
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:
Explanation:
m = Mass of each rod
L = Length of rod = Radius of ring
= Mass of ring
Moment of inertia of a spoke
For 8 spokes
Moment of inertia of ring
Total moment of inertia
The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is .