Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Answer:
a) the velocity of the implant immediately after impact is 20 m/s
b) the average resistance of the implant is 40000 N
Explanation:
a) The impulse momentum is:
mv1 + ∑Imp(1---->2) = mv2
According the exercise:
v1=0
∑Imp(1---->2) = F(t2-t1)
m=0.2 kg
Replacing:
if F=2 kN and t2-t1=2x10^-3 s. Replacing
b) Work and energy in the system is:
T2 - U(2----->3) = T3
where T2 and T3 are the kinetic energy and U(2----->3) is the work.
Replacing:
Answer:
composition of alpha phase is 27% B
Explanation:
given data
mass fractions = 0.5 for both
composition = 57 wt% B-43 wt% A
composition = 87 wt% B-13 wt% A
solution
as by total composition Co = 57 and by beta phase composition Cβ = 87
we use here lever rule that is
Wα = Wβ ...............1
Wα = Wβ = 0.5
now we take here left side of equation
we will get
= 0.5
= 0.5
solve it we get
Ca = 27
so composition of alpha phase is 27% B
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.