Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:Vb=-6i-(-0.1ωab+8)j m/s
Explanation:
Va=V0+Va0
Va=V0+(ra0 x ωao)
ω=Angular velocity of link A0
Using r0a=0.1m;
Va=V0+(0.1i x ω0a K)
Va=0
ixk=j
Va=0+0.1ω0aj
Calculating te velocity of using te equation below
Vb=Va+Vba
Vb=Va+ωab x rba
ωab=40rad/s
rab=-0.21i+0.15j
Va=0.1ω0aj
Vb=Va+ωabxrba
Vb=0.1ω0aj+40k x -(0.21i+0.15j)
Vb=0.1ω0aj-8j-6i
Vb=-6i-(-0.1ωab+8)j m/s
Answer:
Explanation:
The line from "On becoming an inventor" that says:
"I found that at college I could get help from my teachers with solving business problems and in learning new techniques for designing new things"
On Becoming an Inventor was by Dean Kamen an American Engineer, Inventor and Businessman
Answer:
The final temperature of water is 381.39 °C.
Explanation:
Given that
Mass of water = 5 kg
Heat transfer at constant pressure Q = 2960 KJ
Initial temperature = 240 °C
We know that heat transfer at constant pressure given as follows
We know that for water
Lets take final temperature of water is T
So
T=381.39 °C
So the final temperature of water is 381.39 °C.