Answer:
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Lois a Meg a
Lois b
Meg
Lois a Meg a Stewie a
Lois a Meg a Stewie a Brian b
Brian Stewie
Meg a
Meg b
Meg
Lois a Meg a
Lois a Meg a Brian b
Brian
Answer:
500 N
Explanation:
Given;
Mass of the car, M = 1000 kg
initial speed of the car, u = 0 m/s
Final speed of the car, v = 60 m/s
Time, t = 1 min = 60 s
Now,
Force, F is given as:
F = Ma
where,
a is the acceleration
From the Newton's equation of motion, we have
v = u + at
on substituting the values, we get
60 = 0 + a × 60
or
a = 1 m/s²
Thus,
Force = 1000 × 1 = 1000 N
now,
this force will be equal to the friction force provided by the rear wheels
let the friction force on a single rear wheel be 'f'
thus,
2f = 1000 N
or
f = 500 N
Answer:
BMEP = 8.904 bar
IMEP = 11.13 bar
Explanation:
given data
area A = 3250 mm²
length L = 73 mm
calibration factor = 0.2 bar/mm
mechanical efficiency = 80 %
to find out
IMEP and the BMEP
solution
first we calculate the BMEP break mean efficiency pressure tat is express as
BMEP =
here A is area and L is length
so BMEP is
BMEP = = 44.52 mm
BMEP = 44.52 × 0.2 bar = 8.904 bar
and
we know mechanical efficiency is
mechanical efficiency =
so put here value we get IMEP
IMEP =
IMEP = 11.13 bar
Answer: True
Explanation:Typical metals have a property of ductility and malleability that is metals can be drawn into wires or any other shape by beating or stretching the metal by putting the tensile strength or shear strength that pulls them apart . But while compression the metals are squeezed together which affects the hardness of a metal and they are not able to bear the compression force well and thus cannot show elastic-plastic behavior while compression .Therefore the statement given is true typical metals show elastic-plastic behavior in tension and shear but not in compression.