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Katyanochek1 [597]
3 years ago
14

Physics question A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0kN from its engines slow

s it down at a rate of 1.20 m/s^2, but if an upward thrust of only 10.0 kN is applied, it speeds up at rate of .80m/s^2.
Physics
1 answer:
Evgesh-ka [11]3 years ago
4 0
 F=m*a and m is constant on any planet 
25000-m*g=m*1.2 
10000-m*g=-m*0.80 
m*g is the weight 
25000/1.2-m*g/1.2=m 
10000/0.80-m*g/0.80=-25000/1.2+m*g/1.2 solve for m*g 
m*g=(10000/0.80+25000/1.2)/ (1/1.2+1/0.80) 
16 kN 
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The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used
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Answer:

h = 4 in

Explanation:

GIVEN DATA:

volume of tin= 16 \pi

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construct formula for surface area

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A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?
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\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}

  • Speed of the mobile = 250 m/s
  • It starts decelerating at a rate of 3 m/s²
  • Time travelled = 45s

\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}

  • Velocity of mobile after 45 seconds

\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}

We can solve the above question using the three equations of motion which are:-

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  • v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}

We are provided with,

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  • t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

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<u>━━━━━━━━━━━━━━━━━━━━</u>

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The vertical pressure of the water is calculated as follows;

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Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

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