Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>
Answer:
Explanation:
given data:
height of tank = 60cm
diameter of tank =40cm
accelration = 4 m/s2
suppose x- axis - direction of motion
z -axis - vertical direction
= water surface angle with horizontal surface
accelration in x direction
accelration in z direction
slope in xz plane is
the maximum height of water surface at mid of inclination is
the maximu height of wwater to avoid spilling is
= 60 - 8.2
the height requird if no spill water is
Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
Answer:
Find the answer in the attached in the order
Option a, Option c and Option b
Explanation:
