Answer:
The engine consists of a fixed cylinder and a moving piston. The expanding combustion gases push the piston, which in turn rotates the crankshaft. Ultimately, through a system of gears in the power-train, this motion drives the vehicle's wheels.
Explanation:
Answer:
![V_{c}=1.396x10^{-28} m^{3} /unit.cell](https://tex.z-dn.net/?f=V_%7Bc%7D%3D1.396x10%5E%7B-28%7D%20%20m%5E%7B3%7D%20%2Funit.cell)
Explanation:
z = number of atoms
M = Molar mass of zirconium
N = Avogadro’s number
Vc = volume of zirconium unit cell
d = density
![z=12x\frac{1}{6}+2x\frac{1}{2}+3=6](https://tex.z-dn.net/?f=z%3D12x%5Cfrac%7B1%7D%7B6%7D%2B2x%5Cfrac%7B1%7D%7B2%7D%2B3%3D6)
z = 6 atoms per unit cell
M = 91.224 g/mol
N = ![6.023x10^{23} atoms/mol](https://tex.z-dn.net/?f=6.023x10%5E%7B23%7D%20%20atoms%2Fmol)
d = ![6.51g/cm^{3}](https://tex.z-dn.net/?f=6.51g%2Fcm%5E%7B3%7D)
![V_{c}=\frac{zxM}{dxN}](https://tex.z-dn.net/?f=V_%7Bc%7D%3D%5Cfrac%7BzxM%7D%7BdxN%7D)
![V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }](https://tex.z-dn.net/?f=V_%7Bc%7D%3D%5Cfrac%7B6x%2891.224g%2Fmol%29%7D%7B%286.51g%2Fcm%5E%7B3%7D%29%20x%286.023x10%5E%7B23%7Datoms%2Fmol%29%20%7D)
![V_{c}=1.396x10^{-22} cm^{3} /unit.cell](https://tex.z-dn.net/?f=V_%7Bc%7D%3D1.396x10%5E%7B-22%7D%20%20cm%5E%7B3%7D%20%2Funit.cell)
![V_{c}=1.396x10^{-28} m^{3} /unit.cell](https://tex.z-dn.net/?f=V_%7Bc%7D%3D1.396x10%5E%7B-28%7D%20%20m%5E%7B3%7D%20%2Funit.cell)
Answer:
6.99 x 10⁻³ m³ / s
Explanation:
Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.
h = difference in the height of water column at two ends of delivery pipe
6 - 1 = 5 m
Velocity of flow of water
v = √2gh
= √ (2 x 9.8 x 5)
= 9.9 m /s
Volume of water flowing per unit time
velocity x cross sectional area
= 9.9 x 3.14 x .015²
= 6.99 x 10⁻³ m³ / s
Answer:
135 hour
Explanation:
It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.
We have to find the time necessary to achieve the same concentration at a 6 mm position.
we know that
where x is distance and t is time .As the temperature is constant so D will be also constant
So
then
we have given
and we have to find
putting all these value in equation
![\frac{2^2}{15}=\frac{6^2}{t_2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5E2%7D%7B15%7D%3D%5Cfrac%7B6%5E2%7D%7Bt_2%7D)
so
A. I believe, lmk if I’m right