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bixtya [17]
2 years ago
14

Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.

Physics
1 answer:
LiRa [457]2 years ago
7 0

Answer:

         Time - taken = 2.5 s

          deceleration= -8 m/s²

Solution:

            Given:

                     speed, v = 8 m/s

                 distance, d = 20m

                     

              To Find:

                     deacceleration = ?

               

               As we know speed is defined as

                          v = d/t

                plugging in the values

                          t =  20/ 8

                          t = 2.5s

                Now from deceleration formula

                        a =  - v/ t

                        a = - 20/ 2.5

                        a = - 8 m/s²

          Thus, the time taken and acceleration is 2.5 s and -8 m/s²

          respectively.

          Learn more about deceleration here:

                brainly.com/question/13354629

                       #SPJ4  

               

                       

             

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Answer:

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Explanation:

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The final momentum is the motion before the action so:

a).

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2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

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Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

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Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

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3 years ago
Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t
ruslelena [56]

Answer:

x_1= 0.0425m

Explanation:

Using the tension in the spring and the force of the tension can by describe by

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Answer:

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3 years ago
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