Answer:
Explanation:
Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to 
, where m is the mass and M the molar mass of the gas, and the density is 
.
For air 
and 
So, 
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:
HIGH from the supply voltage
LOW from ground
Explanation:
The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high, BUT must be verified), and the LOW signal from ground.
If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range. Then place the probe in the ground of the circuit (must be a big copper area). Finally leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.
Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.
