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:
a) temperature: random error
b) parallax: systematic error
c) using incorrect value: systematic error
Explanation:
Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.
Answer:
375 KPa
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 125 KPa
Initial temperature (T₁) = 300 K
Final temperature (T₂) = 900 K
Final pressure (P₂) =?
The new (i.e final) pressure of the gas can be obtained as follow:
P₁/T₁ = P₂/T₂
125 / 300 = P₂ / 900
Cross multiply
300 × P₂ = 125 × 900
300 × P₂ = 112500
Divide both side by 300
P₂ = 112500 / 300
P₂ = 375 KPa
Thus, the new pressure of the gas is 375 KPa
Answer:
I do!!
Explanation:
I have to sit for 3 hours lol♀️
