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natta225 [31]
2 years ago
14

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.

--------------------

<em>I hope this helps, and I apologize if I get the answer wrong for you, I am 100 percent hopeful that my answer is correct. However if it isn't, please tell me and I will make up for my mistakes. Thank you and have a great rest of your day or night! :></em>

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Tranquilizing drugs that inhibit sympathetic nervous system activity often effectively reduce people's subjective experience of
bagirrra123 [75]

Answer: There are different types of emotion theories. The one we will use here to explain is the two-factor theory otherwise known as Schachter-Singer theory. It's the best way to look at what is going on with how tranquilliser affect emotion.

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However, the cannon-Bard theory suggest arousal and emotions occurred the same time, which makes it difficult in explaining emotion-reducing effect.

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3 years ago
'Energy' has the potential to:
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answer:

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2 years ago
#198. Moment of inertia about center of a segmented bar A bar of width is formed of three uniform segments with lengths and area
zaharov [31]

Complete Complete

The complete question is shown on the first uploaded image

Answer:

The moment of inertia of the bar about the center of mass is

I_r = 1888.80  \  kg m^2

Explanation:

The free body diagram  is shown on the second uploaded image

From the diagram we see that is

The mass of each segment is

          m_1 = \rho_1  l_1 w = 1 * 6 * 2 = 12

          m_1 = \rho_2  l_2 w = 8 * 6 * 2 = 96

          m_1 = \rho_2  l_2 w = 5 * 5 * 2 = 50

The distance from the origin to the center of the segments i.e the center of masses for the individual segments

   x_2 = \frac{6}{2} + 6 = 9 m

   x_3 = \frac{4}{2} + 12 = 14 m

           

The  resultant center of mass is mathematically evaluated as

              x_r = \frac{m_1 * x_1 + m_2 *x_2 + m_3 *x_3}{m_1 + m_2 + m_3}    

        =   \frac{12 * 3 + 96 *9 + 50 *14}{12+ 96 + 50}

                      x_r = 10.13m        

The moment of Inertia of each segment of the bar is mathematically evaluated

             I_1 =\frac{m_1}{12}(l_1^2 + w^2) =    \frac{12}{12}(1^2 + 2^2)        

                   I_1 = 4 \ kgm^2

             I_2 =\frac{m_2}{12}(l_2^2 + w^2)  =    \frac{96}{12}(6^2 + 2^2)

                 I_2 = 320 \ kgm^2

             I_3 =\frac{m_3}{12}(l_3^2 + w^2)  =    \frac{50}{12}(4^2 + 2^2)        

                   I_2 = 83.334 \ kgm^2        

According to parallel axis theorem the moment of inertia about the center (x_r) is mathematically evaluated as

           I_r = (I_1 + m_1 r_1^2) + (I_2 + m_2 r_2^2) +(I_3 + m_3 r_3^2)

   I_r = (I_1 + m_1 |x_r - x_1|^2) + (I_2 + m_2 |x_r - x_2|^2) +(I_3 + m_3 |x_r - x_3|^2)

   I_r = (4  + 12 |10.13 - 3|^2) + (320 + 96 |10.13 - 9|^2) +(83.334 + 50 |10.13 - 14|^2)        

      I_r = 1888.80  \  kg m^2

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(4) Chief Information Officers (CIOs), IT security program managers, auditors, and other security-oriented personnel (e.g., system and network administrators, and system/application security officers) must receive training in information security basics and broad training in security planning, system and application security management, system/application life cycle management, risk management, and contingency planning.

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3 years ago
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A Wii remote flung from a hand through a TV, with a kinetic energy of 1.44J and a mass of 4.5kg. Whats the velocity?
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Answer:

0.8

Explanation:

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