Answer:
Kindly check explanation
Explanation:
(a) μMN, (b) N/μm, (c) MN/ks2, and (d) kN/ms.
(a) μMN = (10^-6 * 10^6) = 10^(-6 + 6) = 10^0 = N
b) N/μm = N / 10^-6 m = 10^6 * N/m = MN/m
(c) MN/ks2 = 10^6N / (10^3 s)^2
10^6 N / 10^6s^2 = 10^6 * 10^-6 N /s^2 = N/s^2
D) kN/ms = 10^3N / 10^-3 s = 10^3 * 10^3 * N/s = 10^6N / s = MN/s
2)
a) 0.000431kg = 431 × 10^6 kg = 431 * 10^9g = 431Gg
b) 35.3 × 10^3 N
10^3 = kilo(K)
35.3 KN
C) 0.00532km = 5.32 * 10^3 km = 5.32 * 10^3 * 10^3 = 5.32 * 10^6 = 5.32Mm
3) Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/nm, and (c) mN/(kg⋅μs).
a) Mg/ms = 10^6g / 10^-3s = 10^6 * 10^3 g/s = 10^9 g/s = Gg/s
b) N/nm = N / 10^-9 m = 10^9 N/m = GN/m
c) mN/(kg⋅μs) = 10^6N / kg(10^-6s) = 10^12N/(kg.s)
= TN/(kg.s)
Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus 
From the refrigerant table A11-A13

sat vapor
m=

b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:the fractional conversion of methane is 12.5%
Explanation:The reaction represent the combustion of methane to produce Co2 and steam.
CH4 +2O2_CO2 + 2H2O
From gay lussac law of proportionality
1mol of CH4 requires 2mol of Oxygen to produce 1 mol of CO2 and 2mol of H2O
So from the combining ratio,25% of O2 will fractional produce 25×1/2% of CH4.
While 12.5% of CO2 and 25% of steam is also produced .so in essence 2.5% of CO2 was lost in the reaction.
Answer:
net force acting on the floor is 100 kN
Explanation:
Given data:


dimension of floor = 2 m \times 0.5 m
we know that
Net force can be calculated as follow




Therefore net force acting on the floor is 100 kN