Answer:
The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.
Answer:
a)
, b)
,
,
, c)
,
,
, ![\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmean%7D%20%3D%20%5Cfrac%7B1%20%2B%20%5Csqrt%7B2%7D%20%7D%7B6000%7D%5C%2Ch)
Explanation:
a) The total number of users that can be accomodated in the system is:
![n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B10%5C%2Ckm%5E%7B2%7D%7D%7B1%5C%2C%5Cfrac%7Bkm%5E%7B2%7D%7D%7Bcell%7D%20%7D%5Ccdot%20%28100%5C%2C%5Cfrac%7Busers%7D%7Bcell%7D%20%29)
![n = 1000\,users](https://tex.z-dn.net/?f=n%20%3D%201000%5C%2Cusers)
b) The length of the side of each cell is:
![l = \sqrt{1\,km^{2}}](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B1%5C%2Ckm%5E%7B2%7D%7D)
![l = 1\,km](https://tex.z-dn.net/?f=l%20%3D%201%5C%2Ckm)
Minimum time for traversing a cell is:
![\Delta t_{min} = \frac{l}{v}](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmin%7D%20%3D%20%5Cfrac%7Bl%7D%7Bv%7D)
![\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmin%7D%20%3D%20%5Cfrac%7B1%5C%2Ckm%7D%7B30%5C%2C%5Cfrac%7Bkm%7D%7Bh%7D%20%7D)
![\Delta t_{min} = \frac{1}{30}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmin%7D%20%3D%20%5Cfrac%7B1%7D%7B30%7D%5C%2Ch)
The maximum time for traversing a cell is:
![\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmax%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B2%7D%5Ccdot%20l%20%7D%7Bv%7D)
![\Delta t_{max} = \frac{\sqrt{2} }{30}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmax%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B30%7D%5C%2Ch)
The approximate time is giving by the average of minimum and maximum times:
![\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmean%7D%20%3D%20%5Cfrac%7B1%2B%5Csqrt%7B2%7D%20%7D%7B2%7D%5Ccdot%5Cfrac%7Bl%7D%7Bv%7D)
![\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmean%7D%20%3D%20%5Cfrac%7B1%20%2B%20%5Csqrt%7B2%7D%20%7D%7B60%7D%5C%2Ch)
c) The total number of users that can be accomodated in the system is:
![n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B10%5Ctimes%2010%5E%7B6%7D%5C%2Cm%5E%7B2%7D%7D%7B100%5C%2Cm%5E%7B2%7D%7D%5Ccdot%20%28100%5C%2C%5Cfrac%7Busers%7D%7Bcell%7D%20%29)
![n = 10000000\,users](https://tex.z-dn.net/?f=n%20%3D%2010000000%5C%2Cusers)
The length of each side of the cell is:
![l = \sqrt{100\,m^{2}}](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B100%5C%2Cm%5E%7B2%7D%7D)
![l = 10\,m](https://tex.z-dn.net/?f=l%20%3D%2010%5C%2Cm)
Minimum time for traversing a cell is:
![\Delta t_{min} = \frac{l}{v}](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmin%7D%20%3D%20%5Cfrac%7Bl%7D%7Bv%7D)
![\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmin%7D%20%3D%20%5Cfrac%7B0.01%5C%2Ckm%7D%7B30%5C%2C%5Cfrac%7Bkm%7D%7Bh%7D%20%7D)
![\Delta t_{min} = \frac{1}{3000}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmin%7D%20%3D%20%5Cfrac%7B1%7D%7B3000%7D%5C%2Ch)
The maximum time for traversing a cell is:
![\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmax%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B2%7D%5Ccdot%20l%20%7D%7Bv%7D)
![\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmax%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B3000%7D%5C%2Ch)
The approximate time is giving by the average of minimum and maximum times:
![\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmean%7D%20%3D%20%5Cfrac%7B1%2B%5Csqrt%7B2%7D%20%7D%7B2%7D%5Ccdot%5Cfrac%7Bl%7D%7Bv%7D)
![\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h](https://tex.z-dn.net/?f=%5CDelta%20t_%7Bmean%7D%20%3D%20%5Cfrac%7B1%20%2B%20%5Csqrt%7B2%7D%20%7D%7B6000%7D%5C%2Ch)
Answer: 3 different types of people using Highway Transportation system are :
1. Pedestrians - should be given priority while driving.
2. Cyclists - should be given enough space on road.
3. Motorcyclists- should be given enough space on road.
Explanation:
1. Pedestrians - Most of pedestrians use to walk on the footpath along side road that keeps them on a safe side. But there are places without footpath along side road, in that case they have to walk on the road itself. Here, we need to take care for them. We need to wait in case they are crossing road and also check for them while taking a turn.
2. Cyclists - They travel on road but are tough to figure out. They travel at a slower pace compared to cars. To avoid any accident with them, we are supposed to give them enough space which should be equivalent to a car's space.
3. Motorcyclists - They can pass by very closely and also come between lanes. Most of the things to be considered here are same as that of cyclists. Here also, we need to check for them carefully while taking a turn. Also, need to give them enough space.
1. A base pair is a fundamental unit of double-stranded nucleic acids consisting of two nucleobases bound to each other by hydrogen bonds.
2.Proteins
3.DNA is made up of molecules called nucleotides. Each nucleotide contains a phosphate group, a sugar group and a nitrogen base. The four types of nitrogen bases are adenine (A), thymine (T), guanine (G) and cytosine (C). The order of these bases is what determines DNA's instructions, or genetic code.
4. any of a group of basic proteins found in chromatin.
5.46 chromosomes
Hope i helped x
Answer:
Vc2= V(l+e) ^2/4
Vg2= V(l-e^2)/4
Explanation:
Conservation momentum, when ball A strikes Ball B
Where,
M= Mass
V= Velocity
Ma(VA)1+ Mg(Vg)2= Ma(Va)2+ Ma(Vg)2
MV + 0= MVg2
Coefficient of restitution =
e= (Vg)2- (Va)2/(Va)1- (Vg)1
e= (Vg)2- (Va)2/ V-0
Solving equation 1 and 2 yield
(Va)2= V(l-e) /2
(Vg)2= V(l+e)/2
Conservative momentum when ball b strikes c
Mg(Vg)2+Mc(Vc)1 = Mg(Vg)3+Mc(Vc)2
=> M[V(l+e) /2] + 0 = M(Vg)3 + M(Vc) 2
Coefficient of Restitution,
e= (Vc)2 - (Vg)2/(Vg)2- (Vc)1
=> e= (Vc)2 - (Vg)2/V(l+e) /2
Solving equation 3 and 4,
Vc2= V(l+e) ^2/4
Vg2= V(l-e^2)/4