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Free_Kalibri [48]
2 years ago
15

The KEOM tower across the street from West is approx. 155 meters high

Physics
1 answer:
blondinia [14]2 years ago
7 0

You told your insane friend that it will take the book 5.6 s to get to the ground.

From the question given above, the following data were obtained:

Height (H) = 155 m

<h3>Time (t) =? </h3>

<u>NOTE</u>: Acceleration due to gravity (g) = 10 m/s²

The time taken for the book to get to ground can be obtained as follow:

<h3>H = ½gt²</h3>

155 = ½ × 10 × t²

155 = 5 × t²

<h3>Divide both side by 5</h3>

t^{2}  = \frac{155}{5}\\\\

t² = 31

<h3>Take the square root of both side </h3>

t = \sqrt{31}

<h3>t = 5.6 s</h3>

Thus, the time taken for the book to get to the ground is 5.6 s

Hence, you told your insane friend that it will take the book 5.6 s to get to the ground.

Learn more: brainly.com/question/24903556

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The term "aerobic" refers to the body's ability to provide energy through the use of oxygen.  The term anaerobic refers to the body's ability to provide energy without the use of oxygen.

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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
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Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

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       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

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what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

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                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

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                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

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                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

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