Solution:
Given:
Calcium and Oxygen ions each having valence = 2 or 2 valence electrons
seperation distance, r = 1.25nm = m
Coulombian force for two point charges seperated by a distance 'r' is given by:
F = (1)
Now, we know that
q = ne
where, n = no. of electrons
e = charge of an electron = C
=
e = 2e
=
e = 2e
Using Eqn (1)
F =
F = N
Since, the force is between two opposite charged ions, it is attractive in nature.
Answer:
a. 121 Btu/lb
b. 211.8lb
c. 2.69/pc
Explanation:
See the attachments please
Solution :
cs=zeros(9001);
ca=zeros(9001);
cp=zeros(9001);
psi=zeros(9001);
t=[0:0.1:900];
cs(1)=0.5;
ce(1)=0.001;
cp(1)=0;
ca(1)=0;
psi(1)=0;
for i=1:1:9000
cs(i+1)=cs(i)-0.1*((0.015*cs(i))/(5.53+cs(i)));
cp(i+1)=cp(i)+0.1*((0.015*cs(i))/(5.53+cs(i))-0.0026*cp(i));
ca(i+1)=ca(i)+0.1*0.0026*cp(i);
psi(i+1)=((cp(i+1)-cp(i)))/((cs(i)-cs(i+1)));
end
plot(t,cs,t,cp,t,ca);
plot(t,psi);
The radius of the specimen is 60 mm
<u>Explanation:</u>
Given-
Length, L = 60 mm
Elongated length, l = 10.8 mm
Load, F = 50,000 N
radius, r = ?
We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this
elongation using Equation:
ε = Δl / l₀
ε = 10.8 / 60
ε = 0.18
We know,
σ = F / A
Where A = πr²
According to the stress-strain curve of brass alloy,
σ = 440 MPa
Thus,
Therefore, the radius of the specimen is 60 mm