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
Part a)
As we know that
here we know that
P1 = 20 bar
V1 = 0.5 m^3
V2 = 2.75 m^3
from above equation
so final state pressure will be 2 bar
Part b)
now in order to find the work done
Answer: yes.
Explanation: The light that will be incidented on that metal is visible light.
It depends on 3 factors:
1. The temperature
2. The specific heat capacity of the metal
3. The thermal conductivity of the metal.
The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.
When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer.
Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.
Explanation:
Measure unstretched length of spring, L. E.g. L = 0.60m.
Set mass to a convenient value (e.g. m = 0.5kg).
Hang mass.
Measure new spring length, L'. E.g. L' = 0.70m.
Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m
Use mg = ke (in equilibrium weight = tension)
k = mg/e
Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).
k = 0.5*10/0.10 = 50 N/m
Repeat for a few different masses. (L always stays the same.)
Take the average of your k values.
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
