A semicircular or circular torch movement should be used when
1 answer:
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Answer:
Below see details
Explanation:
A) It is attached. Please see the picture
B) First to calculate the overall mean,
μ=65∗25/75+80∗25/75+95∗25/75
μ=65∗25/75+80∗25/75+95∗25/75 = 80
Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634
And E(MSE) = σ^2= 9
C) Yes, it is substantially large than E(MSE) in this case.
D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑ = mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
(a) dynamic viscosity =
(b) kinematic viscosity =
Explanation:
We have given temperature T = 288.15 K
Density
According to Sutherland's Formula dynamic viscosity is given by
, here
μ = dynamic viscosity in (Pa·s) at input temperature T,
= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
= reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
= 291.15
when T = 288.15 K
For kinematic viscosity :