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3 years ago

15

Consider a normal shock wave in air. The upstream conditions are given by M1=3, p1 = 1 atm, and r1 = 1.23 kg/m3. Calculate the d

ownstream values of p2, T2, r2, M2, u2, p02 and T02. (Anderson 3.4)

1 answer:

mart [117]3 years ago

7 0

Answer and Explanation:

The answer is attached below

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Water flows at low speed through a circular tube with inside diameter of 2 in. A smoothly contoured body of 1.5 in. diameter is

Art [367]

Answer:

Pressure = 11.38 psi

Force = 13.981 Ibf

Explanation:

Step by step solution is in the attached document.

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3 years ago

Answer true or false 3.Individual people decide what will be produced in a command<br> oconomy

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Answer:

False

Explanation:

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2 years ago

The seers were of the opinion that_____ . *

AURORKA [14]

Answer:

✔️a healthy mind resides in a healthy body.

Explanation:

The seers were of the opinion that "a healthy mind resides in a healthy body."

Just like the English translation of a famous quotation from Thales, pre-Socratic Greek philosopher puts it "a sound mind in a sound body"; which tries to demonstrate the close connections that exists in bodily well-being and one's ability to enjoy life.

The seers were actually of the opinion that a healthy mind resides in a healthy body. It implies that there is connection between the body and the mind. When the body catches an illness, the mind and other parts of the body are affected. When our minds are not healthy, it affects the effective functioning of the body.

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2 years ago

10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de

Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span 20.83\E ft

at mid span = 1.23\E ft

shear stress 7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

$\delta = \frac{wa^2b^2}{3EI}$

$= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}$

$=\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}$

= 20.83\E ft

at mid span

$\delta = \frac{wl^3}{48EI}$

$= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}$

$\delta = 1.23\E ft$

shear stress

$\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi$

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3 years ago