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2 years ago

2) What kinds of food can you eat in space?

Engineering

2 answers:

shtirl [24]2 years ago

6 0

Answer:

fruits, nuts, peanut butter, chicken, beef, seafood, candy, brownies now if we talking about drinks ,coffee, tea, orange juice, fruit punches and lemonade

Explanation:somehow this all are delicious

rjkz [21]2 years ago

5 0

Hi there, so what I know is:

An astronaut can choose from many types of foods such as fruits, nuts, peanut butter, chicken, beef, seafood, candy, brownies, etc. Available drinks include coffee, tea, orange juice, fruit punches and lemonade. As on Earth, space food comes in disposable packages.

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

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Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

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The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

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3 years ago

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Answer:

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Explanation:

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If the coefficient of static friction between the block and the platform is μs = 0.4, determine the maximum speed which the bloc

MrRa [10]

Question:

A disc of radius 6m is rotating about its fixed centre with a constant angular velocity 3rad/s ( in the horizontal plane). A block is also rotating with the disc without slipping. If coefficient of friction between the block and disc is 0.4, determine the maximum speed which the block can attain before it begins to slip. Assume the angular motion of the disk is slowly increasing.

Answer:

Maximum Radius = 0.44m

Given

r = Radius = 6m

w = Angular Velocity = 3rad/s

μ = coefficient of friction between the objects = 0.4

There are four forces acting on the objects

1. Centripetal Forces

2. Friction

3. Blocks weight

4. Normal force on the block

The centripetal force acts inward toward the disk center.

The friction hinders the block's motion.

The block's weight acts on the disk

The normal force of the disk acting on the block.

The block's weight is equal to the normal force on the block because the block sits on the desk surface.

Centripetal force, Fc = mw²r

Friction,Fr = μN where N = Normal force = mg

Friction = μmg

The maximum distance rmax that we could place the block would be when Ff < Fc

We assume they are equal to solve for r

Ff = Fc

μmg = mw²r ---- divide through by m

μg = w²r ----- make r the subject of formula

r = μg/w² where g = 9.8m/s²

Plugging in the values

r = 0.4 * 9.8/3²

r = 0.435555555555555

r = 0.44m ----- Approximated

Explanation:

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