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gladu [14]
3 years ago
6

A car starts from a stopped position at a red light. At the end of 30 seconds, its speed is 20 meters per second. What is the ac

celeration of the car?
A. 1.5 m/s
B. 0.7 m/s
C. 0.7 m/s 2
NEED help please HEELLPP
Physics
1 answer:
Gnoma [55]3 years ago
5 0
A=deltav/t
Vf=20m/s
Vi=0m/s
(20-0)/30
=0.667 m/s^2
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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:

Given:-

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

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

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                   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:

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                  Fac = G*ma*mc / rac^2

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

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       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

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

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

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

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                  Fc = vector Fac + vector Fbc

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

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