Answer:
![Activation\ Energy=2.5\times 10^{-19}\ J](https://tex.z-dn.net/?f=Activation%5C%20Energy%3D2.5%5Ctimes%2010%5E%7B-19%7D%5C%20J)
Explanation:
Using the expression shown below as:
![N_v=N\times e^{-\frac {Q_v}{k\times T}](https://tex.z-dn.net/?f=N_v%3DN%5Ctimes%20e%5E%7B-%5Cfrac%20%7BQ_v%7D%7Bk%5Ctimes%20T%7D)
Where,
is the number of vacancies
N is the number of defective sites
k is Boltzmann's constant = ![1.38\times 10^{-23}\ J/K](https://tex.z-dn.net/?f=1.38%5Ctimes%2010%5E%7B-23%7D%5C%20J%2FK)
is the activation energy
T is the temperature
Given that:
![N_v=2.3\times 10^{13}](https://tex.z-dn.net/?f=N_v%3D2.3%5Ctimes%2010%5E%7B13%7D)
N = 10 moles
1 mole = ![6.023\times 10^{23}](https://tex.z-dn.net/?f=6.023%5Ctimes%2010%5E%7B23%7D)
So,
N = ![10\times 6.023\times 10^{23}=6.023\times 10^{24}](https://tex.z-dn.net/?f=10%5Ctimes%206.023%5Ctimes%2010%5E%7B23%7D%3D6.023%5Ctimes%2010%5E%7B24%7D)
Temperature = 425°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (425 + 273.15) K = 698.15 K
T = 698.15 K
Applying the values as:
![2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}](https://tex.z-dn.net/?f=2.3%5Ctimes%2010%5E%7B13%7D%3D6.023%5Ctimes%2010%5E%7B24%7D%5Ctimes%20e%5E%7B-%5Cfrac%20%7BQ_v%7D%7B1.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%20698.15%7D)
![ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}](https://tex.z-dn.net/?f=ln%5B%5Cfrac%20%7B2.3%7D%7B6.023%7D%5Ctimes%2010%5E%7B-11%7D%5D%3D-%5Cfrac%20%7BQ_v%7D%7B1.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%20698.15%7D)
![Q_v=2.5\times 10^{-19}\ J](https://tex.z-dn.net/?f=Q_v%3D2.5%5Ctimes%2010%5E%7B-19%7D%5C%20J)
Answer:
9500 kJ; 9000 Btu
Explanation:
Data:
m = 100 lb
T₁ = 25 °C
T₂ = 75 °C
Calculations:
1. Energy in kilojoules
ΔT = 75 °C - 25 °C = 50 °C = 50 K
![m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}](https://tex.z-dn.net/?f=m%20%3D%20%5Ctext%7B100%20lb%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20kg%7D%7D%7B%5Ctext%7B2.205%20lb%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1000%20g%7D%7D%7B%5Ctext%7B1%20kg%7D%7D%3D%204.54%20%5Ctimes%2010%5E%7B4%7D%5Ctext%7B%20g%7D%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Brcl%7Dq%20%26%20%3D%20%26%20mC_%7B%5Ctext%7Bp%7D%7D%5CDelta%20T%5C%5C%26%20%3D%20%26%204.54%20%5Ctimes%2010%5E%7B4%7D%5Ctext%7B%20g%7D%20%5Ctimes%204.18%20%5Ctext%7B%20J%24%5Ccdot%24K%24%5E%7B-1%7D%24g%24%5E%7B-1%7D%24%7D%20%5Ctimes%2050%20%5Ctext%7B%20K%7D%5C%5C%20%26%20%3D%20%26%209.5%20%5Ctimes%2010%5E%7B6%7D%5Ctext%7B%20J%7D%5C%5C%20%26%20%3D%20%26%20%5Ctextbf%7B9500%20kJ%7D%5C%5C%5Cend%7Barray%7D)
2. Energy in British thermal units
![\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}](https://tex.z-dn.net/?f=%5Ctext%7BEnergy%7D%20%3D%20%5Ctext%7B9500%20kJ%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20Btu%7D%7D%7B%5Ctext%7B1.055%20kJ%7D%7D%20%3D%20%5Ctext%7B9000%20Btu%7D)
Technician is correct sorry if im wronghg
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
Its 0.001
0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m