Answer: precision
Explanation: because accuracy is right on there and precision is getting closer and closer
<h2>Answer:</h2>
<h3>Required Answer is as follows :-</h3>
- 3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²
- ¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²
- ¾ = m₁ × (V)²/m₂ × (V²/4)
<h3>Now,</h3>
- ∆m = m₂ - m₁
- ∆m = 16m₂/3 - m₁ = 13m₁/3
- Ratio = (13m₁/3)/ m₁
- Ratio = 13/3
Ratio = 13:3
<h3>Know More :-</h3>
Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg.
What do materials science and engineering graduates do?
Almost two-thirds of materials science graduates are in employment six months after graduation.
The skills developed during a materials science degree allow graduates enter a range of sectors, including working as engineering professionals and in design and marketing roles.
Destination Percentage
Employed 60.4
Further study 24.5
Working and studying 5.1
Unemployed 4
Other 6.1
Graduate destinations for materials science and engineering
Type of work Percentage
Engineering and building 23.9
Marketing, PR and sales 11.9
Business, HR and financial 11.7
Technicians and other professionals 10
Other 42.5
Answer:
Computation of the load is not possible because E(test) >E(yield)
Explanation:
We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if
E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:
Calculation of E(test is as follows)
E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm
=0.0262
Computation of E(yield) is given below:
E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027
Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).
