Answer:
43248 newtons.
Explanation:
Force = mass x accelerations and units of force are newtons which are given in the question.
here mass = 125 of air and 2.2 of fuel, total = 125+2.2=127.5kg/s and the velocity of the exhaust is 340m/s.
force = 340m/s * 127.5kg/s = 43248 newtons technically this is wrong (observe units) but i will expalin how i have taken acceleration as a velocity here and mass/unit time as simply mass.
see force is mass times acceleration or deceleration, here our velocity is not changing therefore it is constant 340m/s but if it were to change and become 0 in one second then there would be -340m/s^2 (note the units ) of deceleration and there would be force associated with it and that force is what i have calculated here. similarly there would be mass in flow rate of mass per second, which is also in that one second of time.
let's calculate error.
error = (actual-calculated)/actual. = (43248-60000)/43248= -38.734% less is ofcourse greater than 2%.
So the load cell is not reading correct to within 2% and it should read 43248newtons.
Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases
where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)=
gas constant ideal for air
2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw
Answer: the shear strength at a depth of 12 ft is 1034.9015 lb/ft²
Explanation:
Given that;
Weight of soil r = 118 lb/ft³
stress parameter C = 250 lb/ft²
φ total = 29°
depth Z = 12 ft
The shear strength on a horizontal plane at a depth of 12ft
ζ = C + δtanφ
where δ = normal stress
normal stress δ = r × z = 118 × 12 = 1416
so
ζ = C + δtanφ
ζ = 250 + 1416(tan29°)
ζ = 250 + 1416(tan29°)
ζ = 250 + 784.9016
ζ = 1034.9015 lb/ft²
Therefore the shear strength at a depth of 12 ft is 1034.9015 lb/ft²
Answer:
A
Hope this helps!
Explanation:
You need to find what the problem is and what can or can not be done to solve that issue/compete the task
