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3 years ago

What color would the sky be if the atmosphere was 100% large molecules and particles like dust and water?

Physics

1 answer:

DedPeter [7]3 years ago

3 0

Answer:

Gray-ish looking like as if the atmosphere were full of only pollution.

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Table C: Average Speeds for Lower Racetrack

Blababa [14]

Answer:

Centripetal Acceleration = v^2/r

= (circumference/time)^2/r

= (2*pi*r/t)²)/r

= ((2³.14*50/14.3)²)/50

= 9.64 m/s²

brainlist?

Explanation:

5 0

3 years ago

"determine the resultant internal loadings acting at the cross sections at points f and g of the frame. set θ = 27º and t = 178

Leviafan [203]

Hi you didn't provide any images to solve the question, hence I am going to solve a different question of same concept so you can have an idea how to tackle such types of questions.(please refer to the attachment for question)

Answer:

<u> Please refer to the attachment for answers and explanation</u>

Explanation:

<u> Please refer to the attachment for answers and explanation</u>

5 0

3 years ago

What two states of matter are pictured in the image below?

Anuta_ua [19.1K]

Answer:

volume liquid

Explanation:

6 0

3 years ago

The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm

DIA [1.3K]

Answer:

$T_{max} = 4.735\,kN\cdot m$

Explanation:

The shear stress due to torque can be calculed by using the following model:

$\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}$

The maximum torque on the section is:

$T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}$

The Torsion Constant for the circular tube is:

$J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})$

$J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]$

$J_{tube} = 4.560\times 10^{-6}\,m^{4}$

Now, the require output is computed:

$T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}$

$T_{max} = 4.735\,kN\cdot m$

7 0

3 years ago

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Find the wavelength of radio waves of frequency 200kHz

Gala2k [10]

$\lambda = \frac{c}{f}$
λ - wavelength, c - the speed of light, f - frequency

$f=200 \ kHz= 200 000 \ Hz \\ \\
\lambda=\frac{300 000 \ [\frac{km}{s}]}{200 000 \ [Hz]}=\frac{3}{2}=1.5 \ [km]$

The wavelength of these waves is 1.5 km.

3 0

3 years ago

Read 2 more answers