Answer:
a) 70.29 %
b) 37%
Explanation:
percent reduction can be found from:
PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2
= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2
=70.29 %
percent elongation can be found from:
EL =L_f - Lo/Lo*100
= (73.17 -53.3/53.3)*100
= 37%
The question is asking whether that statement is true or false. Options are;
A) True
B) False
This is about usage of Swing arm restraints.
<em><u>B) False</u></em>
There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.
- Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.
- This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.
Read more at; brainly.com/question/17972874
Answer:
a) the object floats
b) the object floats
c) the object sinks
Explanation:
when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water
a)
volumen for a cube
V=L^3
L=1.53in=0.0388m
V=0.0388 ^3=5.8691x10^-5m^3=58.69ml
density=m/v
density=13.5g/58.69ml=0.23 g/ml
The wooden block floats because it is less dense than water
b)
m=111mg=0.111g
density=m/v
density=0.111g/0.296ml=0.375g/ml
the metal paperclip floats because it is less dense than water
c)
V=0.93cups=220.0271ml
m=0.88lb=399.1613g
Density=m/v
density=399.1613/220.027ml=1.8141g/ml
the apple sinks because it is denser than water
Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, 
is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for 
while a is taken as 0.003m and Y is already known
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
