Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
<em>4.3 Print count</em>
<em>4.4 count = 0</em>
End the outer loop
4.5 End i loop
End the algorithm
5. End
Answer: new depth will be 3.462m and the water elevation will be 0.462m.
The maximum contraction will be achieved in width 0<w<3
Explanation:detailed calculation and explanation is shown in the image below
Answer:
well you could get some green goblin it disolves all the c rap in sink
Explanation:
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
