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3 years ago

The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'

Physics

1 answer:

blsea [12.9K]3 years ago

8 0

Answer: Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

answer:

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

Determine the no-friction needed speed

First step : calculate the value of R

2πR / 4 = πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

= 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

v = √9.81 * 256 * tan(9.2°) = 20.1 m/s

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The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

From the Fourier's law of conduction:

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

Now calculating the equivalent thermal resistance for conductivity using electrical analogy:

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

For plywood-Styrofoam interface from inside:

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&For Styrofoam-plywood interface from inside:

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

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3 years ago

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Explanation:

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The answer would be B.
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marta [7]

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