Answer:
a) The net force on the ball is instantaneously equal to zero newtons at the top of the flight path.
Explanation:
At an instantenous time at the top of the flight path, the upward force due to the Canon explosion on the ball is just equal to the weight of the ball, this will equate the net force on the ball to zero. At this point the velocity of the ball is zero before it decends down to earth under its own weight.
Answer:
Pressure = 115.6 psia
Explanation:
Given:
v=800ft/s
Air temperature = 10 psia
Air pressure = 20F
Compression pressure ratio = 8
temperature at turbine inlet = 2200F
Conversion:
1 Btu =775.5 ft lbf, 
= 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²
Air standard assumptions:
= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R
k= 1.4
Energy balance:
As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible
hence 
= 20+460 = 480°R
= 533.25°R
Pressure at the inlet of compressor at isentropic condition
= 
= 14.45 psia
Answer:
A
Explanation:
The best method that will yield significantly more accurate result is to use spectrophotometer to read the turbidity of the sample and increase in turbidity is associated with increase biomass.
Answer:
h = 375 KW/m^2K
Explanation:
Given:
Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm
steel thermal conductivity k = 15 W / mK
Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C
Air Temp T_∞ = 100 C
Assuming there are no other energy sources, energy balance equation is:
E_in = E_out
q"_cond = q"_conv
Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2
q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)
=15KW/m^2
Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:
q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )
h = 15000 W / (100 - 60 ) C = 375 KW/m^2K
The answer is c. 4 seconds
