Answer:
Hello there, please check step by step explanations to get answers.
Explanation:
Given that:
5-hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation.
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Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore
and radial component of given velocity is zero
we have
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
so
solvingt for
therefore eccentrcity of orbit is 0.22
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.
