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Helen [10]
2 years ago
15

In a particular application involving airflow over a surface, the boundary layer temperature distribution may be approximated as

T−Ts/T[infinity]−Ts=1−exp⁡(−Pr(u[infinity]y)/????) Where y is the distance normal to the surface and the Prandtl number, Pr=cpμ/k=0.7, is a dimensionless fluid property. If T[infinity]=400K, Ts=300K, and u[infinity]/υ=5000 m-1, what is the surface heat flux? 6.11

Physics
1 answer:
Anni [7]2 years ago
8 0

Answer:

The Surface heat flux is -9205 W/m^2

Explanation:

 Explanation is in the following attachment    

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A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the
kodGreya [7K]

Answer:

3.73994

No,

Explanation:

P_1 = Pressure at the bottom of the lake = 3.5 atm

P_2 = Pressure at the top of the lake = 1 atm

V_b = Volume at the bottom of the lake

V_s = Volume at the top of the lake

T_1 = Temperature at the bottom of the lake = 4 °C

T_2 = Temperature at the top of the lake = 23 °C

From ideal gas law we have the relation

\dfrac{P_1V_b}{T_1}=\frac{P_2V_s}{T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{P_2T_1}{P_1T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{1\times 277.15}{3.5\times 296.15}\\\Rightarrow \dfrac{V_s}{V_b}=0.26738^{-1}\\\Rightarrow \dfrac{V_s}{V_b}=3.73994

The ratio is 3.73994

As Jacques is ascending if he holds his breath his lungs acting like a bubble would expand. Hence, it is not safe to hold his breath while ascending,

8 0
2 years ago
The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t
3241004551 [841]

Answer:

g'=\frac{g__R}{4}

Explanation:

Given:

  • gravitational field strength of moon at a distance R from its center, g__R
  • Distance of the satellite from the center of the moon, h=2R

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

g'=g__{R}} \times \frac{R^2}{(2R)^2}

g'=\frac{g__R}{4}

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0
2 years ago
An astronaut on a space walk floats a little too far away
Lady bird [3.3K]

Answer:

He can throw it away from himself.

Explanation:

Newtons Third Law says that everything has an equal, yet opposite reaction on other objects.

3 0
1 year ago
Read 2 more answers
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame
kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0
1 year ago
Read 2 more answers
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