Answer:
Elastic modulus of steel = 202.27 GPa
Explanation:
given data
long = 110 mm = 0.11 m
cross section 22 mm = 0.022 m
load = 89,000 N
elongation = 0.10 mm = 1 × 
m
solution
we know that Elastic modulus is express as
Elastic modulus = 
................1
here stress is
Stress = 
.................2
Area = (0.022)²
and
Strain = 
.............3
so here put value in equation 1 we get
Elastic modulus = 
Elastic modulus of steel = 202.27 × 
Pa
Elastic modulus of steel = 202.27 GPa
Answer:
d) Binary
Explanation:
Binary search is a search algorithm that finds the position of a value in an ordered array. It compares the value with the element in the middle of the array, if they are not equal, half in which the value cannot be eliminated and the search continues in the remaining half until the value is found.
Answer: Advanced technologixal machines
Explanation: such as big cranes, multiple workers helping creat said structure, and big bull dozers
Answer:
L = 475.718
T = 240.89 ft
M = 23.0195
LC = 472.728
R = 1225 ft
Explanation:
See the attached file for the calculation.
Answer:
Explanation:
We are given:
m = 1.06Kg
T = 22kj
Therefore we need to find coefficient performance or the cycle
= 5
For the amount of heat absorbed:
= 5 × 22 = 110KJ
For the amount of heat rejected:
= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;
= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at 
= 12.4°c we have 442KPa
