5/2055 classes displayed there’s Nooooob changes
Answer:
The growth of crack formation in a corrosive environment.
Explanation:
Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
<u>Original Length = 235.6 mm</u>
Answer: a. 0.4667
b. 0.4667 and C 0.0667
Explanation:
Given Data:
N = population size (10)
n = random selection (2)
r = number of observations = 7
Therefore
f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )
When y = 1
f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 2
f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 0
f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )
= 1 / 15
= 0.0667
Answer:
The constant here is the study outline
Explanation:
In scientific research, the constant variable is that part/variable of the experiment that does not change or is set not to change. Examples include temperature, environment or height.
Assuming the scenery described in this question is an experiment. All the groups presented are bound by a constant during the experiment. The constant here is the study outline. The study outline provided to the students is not going to change.
NOTE: There could be confusion as regards the answer being the final exam grade but that will be the dependent variable as that will be the outcome of the experiment while the time spent to study will be the independent variable.
