Question
__ 16. A bar mensures 300cmtimes 200cm X Im. The specific gravity substance is 8.92 what is the volume of substance? A. 6m^3 B. 60m^3 C. 0.6m^3 D. none 17. According to the question 16^th __ __ A. 8.92times 10^3kg B. 53.52times 10^3kg C. 6kg D. 6times 10^3kg __ 18. Which is unitless expression? A. strain B. Stress C. young modulus D. all __ 19. A wire is stretched 33mm by a force of 150N The force that required to stretch by 5mm is __ A.2500N B.250N C 25N D. 500N __ 20. Which formula indicate the young modulus? A. (stress)/(strain) B. (FL_(D))/(pi gamma 2Delta L) C. (FL_(0))/(Delta LA) D. All iv. Bones 21. Asteel wire of length 6m and diameter of 0.6mm extend by aforce of 60N ,the wire extended by 3mm.calculate a/the applied stress?
Solution
3.9
(242 Votos)
Alexandre
Profissional · Tutor por 6 anos
Resposta
Let's go through each question step by step:**Question 16:**Given:- Dimensions of the bar:
- Specific gravity of the substance: 8.92To find the volume of the substance, we first need to calculate the volume of the bar and then use the specific gravity to find the actual volume of the substance.1. Convert all dimensions to meters: -
-
-
2. Calculate the volume of the bar:
3. The specific gravity (SG) is the ratio of the density of the substance to the density of water. Since the density of water is
, the density of the substance is:
4. The volume of the substance is:
So, the correct answer is:**C.
****Question 17:**Given the volume of the substance from question 16, we can find the mass using the density:
So, the correct answer is:**C. 6 kg****Question 18:**The unitless expressions are:- Strain- Young's modulusSo, the correct answer is:**A. Strain****Question 19:**Given:- Original stretch: 33 mm- Force: 150 NWe need to find the force required to stretch by 5 mm.Using Hooke's Law:
Where
is the spring constant and
is the stretch.First, find
:
Now, find the force for a 5 mm stretch:
So, the correct answer is:**B. 250 N****Question 20:**The formula for Young's modulus is:
So, the correct answer is:**A.
****Question 21:**Given:- Length of the wire: 6 m- Diameter of the wire: 0.6 mm- Force applied: 60 N- Extension: 3 mmWe need to calculate the applied stress.First, convert the diameter to meters:
Calculate the cross-sectional area
:\[A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.0006}{