Answer:
A.) Find the answer in the explanation
B.) Ua = 7.33 m/s , Vb = 7.73 m/s
C.) Impulse = 17.6 Ns
D.) 49%
Explanation:
Let Ua = initial velocity of the rod A
Ub = initial velocity of the rod B
Va = final velocity of the rod A
Vb = final velocity of the rod B
Ma = mass of rod A
Mb = mass of rod B
Given that
Ma = 2kg
Mb = 1kg
Ub = 3 m/s
Va = 0
e = restitution coefficient = 0.65
The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.
Please find the attached files for the solution
Answer:
A. N type impurities
B. P type impurities
Explanation:
A. The impurities contribute free electrons and changing the conducting property of the semi conductor. When a pentavalent impurities in a semi conductor( impurities with five valence electron) , the impurity atom replace some of the semi conductor atoms in the crystal structure where 4 of the valence electron would be involved in bonding of 4 neighbouring semiconductor while leaving the fifth electron to be free(negative charge carrier) which is available for detachment.
B. When a trivalence impurity is added to semiconductor, instead of excess electron, there will be excess hole created by crystals. Reason for this attribute is the trivalence atom will replace some tetra valence semiconductor atom, when three valence electrons of the 3 valence electrons of the trivalent impurity atom make bond with three neighbouring semiconductor which gives rise to lack of electron in the bond of the fourth neighbouring semiconductor which contribute a whole to the crystalline since trivalent impurity contribute excess holes to the crystal of semi conductor, this holes can accept electrons.
Answer:
28,8 m/s
Explanation:
In a steady flow system we can say that m1=m2 which means that the mass flow in the entrance in the same in the outlet. m is flow (kg/s)
we know that 
where V (m/s) is velocity, A (m^2) ia area and v is specific volume (m^3/kg)
Since m1=m2 we can say
clearing the equation
we can specific volume (m^3/kg) from thermodynamic tables
for the entrance is 400°C and 4 MPa is superheated steam and v is : 0,7343 m^3/kg
In the outlet we have saturated vapor with quality (x) of 80%. In this case we get the specific saturated volume for the liquid (vf) and the specific volume for the saturated (vg) gas from the thermodynamic tables. we use the next equation to get (v) for the condition of interest, in this case 80% quality.
v= vf +x*(vg - vf)
where:
x: quality
vf = liquid-saturated-specific-volume
vg =steam-saturated-specific-volume.
for this problem
x = 0,8
vf = 0,00102991
vg = 3,24015
so
we get = 2,593 m^3/kg
The area is the one for a circle
r1 = 0,1 m^2 for area 1
r2=0,5 m^2 for area 2
A1 = 0,0314 m^2
A2 = 0,7853 m^2
we know that V1 is 20 m/s
replacing these values in the equation
we get V2 = 28,2 m/s.
