An angle is observed repeatedly using the same equipment and procedures producing the data below:32 ∘ 37' 00", 32 ∘ 37' 10", 32
1 answer:
Answer:
a) x_average = 34.51077°, b) σ = 2.69°, c) σ = 2.5107°
Explanation:
a) The most likely value of a measure is
x_average = ∑ / n-1
The angle measurements are in the Sixty Base System, to simplify the calculations let's reduce the measurements
1° = 60 min
1 min = 60 s
Table 1
Angle angle converted (x-x_average) (x-x_average)²
32 32,000 2,51077 6.3040
37 00” 37,000 2,48923 6.1963
32 32,000 2,51077 6.3040
37 10” 37.00277 2.492 6.2101
32 32,000 2,51077 6.3040
37 10” 37.1667 2.6559 7.0538
32 32,000 2,51077 6.3040
36 55” 36.9167 2.4059 5.7884
With this values in degree the calculation is easier
x_average = (32 + 37 + 32 + 37.00277 + 32 + 37.1667 + 32 + 36.9167) / 8
x_average = 276.08617 / 8
x_average = 34.51077
b) We look for the standard deviation
σ = √ ( -x_average)² / n-1
σ = √ 50.4646 / (8-1)
σ = 2.69
c) the deviation from the mean
σ = | x -| / n
σ = 20.08611 / 8
σ = 2.5107
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Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
Answer:
1. Graph C
2. Friction
Explanation:
1. The line on all of the graphs shown represents velocity. The formula for velocity is where d is distance and t is time. Focusing on the first lap, the starting point on the graph should be the origin and the "ending" point should be (20, 3). These requirements eliminate graph A as an answer because its "end" is not (20, 3). During the break, the student does not move, so the slope of the line should be completely horizontal. The break lasted for 5 minutes, so the correct graph should have a horizontal line between the points (20, 3) and (25, 3). This requirement eliminates graph B and D because their break is either not long enough (B) or too long (D).
2. Friction slows down the movement of objects. When an object is rough, it produces more friction which causes the object to be slowed more. When an object is smooth, friction slows it less than it would for a rough object.