Yes
Acceleration, in physics, is the rate of change of velocity of an object.
sorry I'm late but the answer is the 2nd option
Answer:
longitudinal engineering strain = 624.16
true strain is 6.44
Explanation:
given data
diameter d1 = 0.5 mm
diameter d2 = 25 mm
to find out
longitudinal engineering and true strains
solution
we know both the volume is same
so
volume 1 = volume 2
A×L(1) = A×L(2)
( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)
( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)
0.1963 ×L(1) = 122.71 ×L(2)
L(1) / L(2) = 122.71 / 0.1963 = 625.16
and we know longitudinal engineering strain is
longitudinal engineering strain = L(1) / L(2) - 1
longitudinal engineering strain = 625.16 - 1
longitudinal engineering strain = 624.16
and
true strain is
true strain = ln ( L(1) / L(2))
true strain = ln ( 625.16)
true strain is 6.44
Answer:
a) I = 0.0198 kg m²
, b) I = 21.85 kg m²
Explanation:
For this exercise we will use the definition of moment of inertia
I = ∫ r² dm
For body with high symmetry they are tabulated
sphere I = 2/5 m r²
bar with respect to center of mass I = 1/12 m L²
let's calculate the mass of each body
bar
ρ = m / V
m = ρ V
m = ρ l w h
where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm
m = 32820 1 0.008 0.04
m = 10.5 kg
Sphere
M = ρ V
V = 4/3 pi r³
M = rgo 4/3 π r³
give us the density 37800 kg / m³ and the radius of 5 cm
M = 37800 4/3 π 0.05³
M = 19.8 kg
a) asks us for the moment of inertia of the sphere with respect to its center of mass
I = 2/5 M r²
I = 2/5 19.8 0.05²
I = 0.0198 kg m²
b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes
I = I_cm + M d2
where d is the distance from the body to the point of interest
I_cm = 0.0198 kg m²
the distance to the pivot point is
l = length of the bar + radius of the sphere
l = 1 + 0.05 = 1.005 m
I = 0.0198 + 19.8 1.05²
I = 21.85 kg m²
ΒγΕ the answer should be a hot day
