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Vaselesa [24]
2 years ago
9

An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi

th the faux fawn for 0.815 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the care before the collision? What is the magnitude of the average acceleration of the car during the collision?
Physics
1 answer:
hoa [83]2 years ago
4 0

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

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8 0
3 years ago
A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise
lesya692 [45]

Answer:

H = 109.14 cm

Explanation:

given,                                                            

Assume ,                                                            

Total energy be equal to 1 unit                                

Balance of energy after first collision = 0.78 x 1 unit

                                                             = 0.78 unit

Balance after second collision = 0.78 ^2 unit

                                                   = 0.6084 unit

Balance after third collision = 0.78 ^3 unit

                                              = 0.475 unit

height achieved by the third collision will be equal to energy remained                                        

H be the height achieved after 3 collision

0.475 ( m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

6 0
3 years ago
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