Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer:
Explanation:
1. With the operands R0, R1, the program would compute AND operation and ADD operation .
2. The operands could truly be signed 2's complement encoded (i.e Yes) .
3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.
When the result of an operation is smaller in magnitude than the smallest value represented by the data type, then arithmetic underflow will occur.
Answer:
Heat gain of 142 kJ
Explanation:
We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.
Δ
⇒ 
Δ
Therefore, the gas gained heat by an amount of 142 kJ.
Answer:
Explanation:
cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²
= 4.94 x 10⁻⁶ m²
stress = 42 x 9.8 / 4.94 x 10⁻⁶
= 83.32 x 10⁶ N/m²
strain = .002902 / 2.7
= 1.075 x 10⁻³
Young's modulus = stress / strain
= 83.32 x 10⁶ / 1.075 x 10⁻³
= 77.5 x 10⁹ N/m²
