Answer:
Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct
Explanation:
He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.
Answer: The actual tracking weight of a stereo cartridge that is set to track at 3 g on a particular changer can be regarded as a continuous rv X with the following
Explanation:
Answer:
The Employee
Explanation:
Because it is there responsibility
Answer:
a) 
, b) 
, c) 
Explanation:
a) The deceleration experimented by the commuter train in the first 2.5 miles is:
![a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5B%2815%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Ch%7D%7B3600%5C%2Cs%7D%20%29%5D%5E%7B2%7D-%5B%2850%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Ch%7D%7B3600%5C%2Cs%7D%20%29%5D%5E%7B2%7D%7D%7B2%5Ccdot%20%282.5%5C%2Cmi%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%7D)
The time required to travel is:
b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be .
c) The final constant deceleration is:
Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g
