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Bess [88]
3 years ago
5

Please, I need help with this.

Physics
2 answers:
adoni [48]3 years ago
8 0

Answer:

42.72 is the magnitude of the football’s displacement.

solmaris [256]3 years ago
6 0

this is the answer :)

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A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
What is the mass of a large ship that has a momentum of 1.60x10^9 kg*m/s and is moving at a velocity of 10m/s?
Elden [556K]

Answer:

160000000 kg.

Explanation:

p=mv

p=1.6x10^9

v=10m/s

rearrange and substitute:

(1.6x10^9)=m(10)

m=(1.6x10^9)/10

m= 1.6x10^8 kg.

7 0
3 years ago
Describe the ratio of kinetic energy and potential energy the skateboarder will have at each of the following points in her ride
amm1812

Answer:

no answer

Explanation:

7 0
2 years ago
Un joven pelotero llamado Saúl en su primer año de ser firmado en grandes ligas, batea varias veces de cuadrangular, el segundo
Brilliant_brown [7]

Answer:

135-15=120

120÷3=40

40+40+(10)+40+(5)=135

8 0
3 years ago
Read 2 more answers
A suspension bridge with weight uniformly distributed along its length has twin towers that extend 65 meters above the road surf
Fynjy0 [20]

Answer:

16.25 m

Explanation:

we know that the equation pf parabola

y=kx^2

from bellow figure the coordinate of parabola is (600,65) that is y=600 and x=65

putting the the value of y and x in the equation of parabola

65=k600^2

k=0.0001805

now the equation is

y=0.0001805x^2

we have to find the value of y at x=300m

so y=0.0001805\times 300^2

y=16.25 m

4 0
3 years ago
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