Malleable and ductile
non metals like plastic also have other properties but can't be malleable and ductile so they r most valuable metallic properties
Answer:
work done = 48.88 × 
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × J
Answer:
the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Explanation:
Given that;
volume of cut = 25,100 m³
Volume of dry soil fill = 23,300 m³
Weight of the soil will be;
⇒ 93% × 18.3 kN/m³ × 23,300 m³
= 0.93 × 426390 kN 3
= 396,542.7 kN
Optimum moisture content = 12.9 %
Required amount of moisture = (12.9 - 8.3)% = 4.6 %
So,
Weight of water required = 4.6% × 396,542.7 = 18241 kN
Volume of water required = 18241 / 9.81 = 1859 m³
Volume of water required = 1859 kL
Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL
Answer:
The new length of the rod is 182 cm.
Explanation:
Given that a rod that was originally 100-cm-long experiences a strain of 82%, to determine what is the new length of the rod, the following calculation must be performed:
100 x 1.82 = X
182 = X
Therefore, the new length of the rod is 182 cm.
Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
