Answer:
The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V
Explanation:
As the complete question is not given the complete question is found online and is attached herewith.
By applying KCL at node 1
Also
Now applying KVL on loop 1 as indicated in the attached figure
Similarly for loop 2
So the system of equations become
Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as
The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V
Answer:
4.13 MPa
Explanation:
Given
Modulus of elasticity = E = 102 GPa
E = 102 * 10^9
Modulus resilience U = 0.79 * 106Pa
Modulus of Resilience U = y²/2E --- Make y the subject of formula
y² = 2EU
y = √(2EU)
y = √(2 * 102 * 10^9 * 0.79 * 106)
y = √1.708296E13
y = 4133153.759540044
y = 4.13E6
y = 4.13MPa
Answer:
-Differential equation: d²T/dx² = 0
-The boundary conditions are;
1) Heat flux at bottom;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
-KdT(L)/dx = h(T(L) - T(water))
Explanation:
To solve this question, let's work with the following assumptions that we are given;
- Heat transfer is steady and one dimensional
- Thermal conductivity is constant.
- No heat generation exists in the medium
- The top surface which is at x = L will be subjected to convection while the bottom surface which is at x = 0 will be subjected to uniform heat flux.
Will all those assumptions given, the differential equation can be expressed as; d²T/dx² = 0
Now the boundary conditions are;
1) Heat flux at bottom;
q(at x = 0) is;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
q(at x = L):
-KdT(L)/dx = h(T(L) - T(water))
A). The burning of coal became a primary fuel
