Answer:
1500Ω
Explanation:
Given data
voltage = 15 V
total Resistance = 4000Ω
potential drop V = 9.375 V
To find out
R2
Solution
we know R1 +R2 = 4000Ω
So we use here Ohm's law to find out current I
current = voltage / total resistance
I = 15 / 4000 = 3.75 × 
A
Now we apply Kirchhoffs Voltage Law for find out R2
R2 = ( 15 - V ) / current
R2 = ( 15 - 9.375 ) / 3.75 ×
R2 = 1500Ω
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa
Answer:
Resistance of copper = 1.54 * 10^18 Ohms
Explanation:
<u>Given the following data;</u>
Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m
Resistivity, P = 1.7 * 10^8 Ωm
Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m
Radius = 0.000325 m
To find the resistance;
Mathematically, resistance is given by the formula;
Where;
- P is the resistivity of the material.
- L is the length of the material.
- A is the cross-sectional area of the material.
First of all, we would find the cross-sectional area of copper.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.000325)²
Area = 3.142 * 1.05625 × 10^-7
Area = 3.32 × 10^-7 m²
Now, to find the resistance of copper;
<em>Resistance = 1.54 * 10^18 Ohms </em>
Answer:
Smaller impurity atom will nullify some of the compressive strain of a dislocation in a crystal. Because, smaller impurity atoms located near a dislocation creates tensile strain on atoms around it thereby partially nullifying compressive strain at the dislocation.
