Answer:
a) 2319.6 kPa
b) 0.027
c) 287.86 kg/m^3
Explanation:
The pressure is determined from table in the appendix for the given temperature:
P_220=2319.6 kPa
to calculate the quality we need to determine the masses of the vapor and the liquid and for that we need the respective specific volumes which can also be found in table.
m_liq = V_liq/α_liq
= 504.2 kg
m_vap = V_vap/α_vap
= 13.94 kg
q = m_vap/m_liq
= 0.027
Finally, the density is simply calculated as follows:
p = m_tot/V_tot
= 518.14 kg/1.8 m^3
= 287.86 kg/m^3
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = 
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = 
· 
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = × 
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
Vf = specific volume of saturated liquid = 0.0217158 ft^3/lb
Vg = specific volume of saturated steam = 0.430129 ft^3/lb
Explanation:
Given data:
water temperature is given as 287-degree Celcius
we have to find Vf and Vg
Vf = specific volume of saturated liquid
Vg = specific volume of saturated steam
we know that from the saturated steam table we can find these value
therefore for temperature 287-degree Celcius
Vf = specific volume of saturated liquid = 0.0217158 ft^3/lb
Vg = specific volume of saturated steam = 0.430129 ft^3/lb
A SELECT command pulls zero or more rows from one or more database tables or views. SELECT is the most often used data manipulation language (DML) command in most applications. See the statement required below.
<h3>What is the SELECT Statement that gives the above results?</h3>
SELECT VendorName, InvoiceNumber, InvoiceDate,
InvoiceTotal - PaymentTotal - CreditTotal AS Balance
FROM Vendors JOIN Invoices
ON Vendors.VendorID = Invoices.VendorID
WHERE InvoiceTotal - PaymentTotal - CreditTotal > 0
ORDER BY VendorName;
Learn more about SELECT Statements at;
brainly.com/question/19338967
#SPJ1
Answer:
Normal force = 0.326N
Explanation:
Given that:
mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg
height of the mass = 1.1 m
radius = 0.2 m
acceleration due to gravity = 9.8 m/s²
We are to determine the normal force pressing on the track at A.
To to that;
Let consider the conservation of energy relation; which says:
mgh = mgr + 1/2 mv²
gh = gr + 1/2 v²
gh - gr = 1/2v²
g(h-r) = 1/2v²
v² = 2g(h-r)
However; the normal force will result to a centripetal force; as such, using the relation
N =mv²/r
replacing the value for v² = 2g(h-r) in the above relation; we have:
Normal force = 2mg(h-r)/r
Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2
Normal force = 0.065268/0.2
Normal force = 0.32634 N
Normal force = 0.326N
