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mina [271]
4 years ago
7

22. (I) A car slows down from 28 m/s to rest in a distance of

Physics
1 answer:
lidiya [134]4 years ago
7 0

Answer:

a = -8.912 m/s²

Explanation:

Given,

The initial velocity of the car, u = 28 m/s

The final velocity of the car, v = 0

The distance traveled by car, d = 88 m

The velocity displacement relation is given by the formula

                                          v = d/t

∴                                         t = d/v

Substituting in the above values in the given equation

                                           t = 88/28

                                            = 3.142 s

The acceleration is given by the formula

                                         a = (v-u)/t

                                            = (0 - 28)/3.142

                                            = -8.912 m/s²

The negative sign is that the car is decelerating.

Hence, acceleration a = -8.912 m/s²

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What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t
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r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

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