Answer:
the police officer cruise each streets precisely once and he enters and exit with the same gate.
Explanation:
NB: kindly check below for the attached picture.
The term ''Euler circuit'' can simply be defined as the graph that shows the edge of K once in a finite way by starting and putting a stop to it at the same vertex.
The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.
Here in this question, the door is the vertex and the road is the edge.
The information needed to detemine a Euler circuit and a Hamilton circuit is;
"the police officer cruise each streets precisely once and he enters and exit with the same gate."
Check attachment for each type of circuit and the differences.
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa
Answer:
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Explanation:
Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.
Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ




τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0
Answer:
6.000006e+15 Or 6.000006000000000000000
Explanation:
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