To solve this problem it is necessary to apply the concepts related to stress failure, stress and last module Young.
Critical stress by definition is given as,
Where,
Strain fracture toughness
Y = Young's module
a = Length surface crack
Our values are given as,
Replacing in our previous equation we have,
Therefore the critical stress is 984.77Mpa
A loess is...
<em>A clastic, silt-sized sediment that is formed by the accumulation of wind-blown dust. 10% of the earth's area is covered by loess or similar deposits. </em>
<em>Hope this helps you to find your answer and if you ever need help with anything else I would be happy to help,</em>
~QueenBeauty666~
maximum allowed value of the speed in roller coaster is given as
now from kinematics we can say
here initial speed will be
acceleration is due to gravity
now we can use this to find the height
so maximum allowed height will be 20.4 m
Answer:
R = 0.237 m
Explanation:
To realize this problem we must calculate the moment of inertia of the wheel formed by a thin circular ring plus the two bars with an axis that passes through its center.
The moments of inertia of the bodies are additive quantities whereby we can add the mounts inertia of the ring and the two bars.
Moment of inertia ring I1 = MR²
Moment of inertia bar I2 = 1/12 ML²
Moment of inertia disk I3 = ½ mR²
Let's calculate the moment of inertia of the wheel
I = I1 + 2 I2
I = MR² + 2 1/12 ML²
The length of the bar is ring diameter
L = 2R
I = 5.65 0.156² + 1/6 9.95 (2 0.156)²
I = 0.1375 + 0.1614
I = 0.2989 kg m²
This is the same moment of inertia of the solid disk,
Disk
I3 = I
I3 = ½ MR²
They give us disk density
ρ = M / V
M = ρ V
M = ρ (pi R² e)
Done is the thickness of the disc, in general it is e= 1 cm = 0.01 m
Let's replace
I3 = ½ ( ρ π R²) R²
I3 = ½ ρ π e R⁴
R⁴ = 2 I3 / ( ρ π e)
R = ( 2 I3 / ( ρ π e)
R⁴ = 2 0.2989 / (5990 π 0.01)
R = 0.237 m
True: Friction depends on the types of surfaces involved and how hard the surfaces push together.