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Vlad [161]
3 years ago
10

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

magnitude of its acceleration in the water?
A. 12 meters/second2
B. 1.5 × 102 meters/second2
C. 78.6 meters/second2
D. 6.5 × 10-3 meters/second2
Physics
2 answers:
lisabon 2012 [21]3 years ago
5 0

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

                            =       zero            - (43 m/s)

                            =          -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

                                          =  (-43 / 0.28)  (m/sec) / sec

                                          =        153.57...  m/s²

                                          =        1.5...  x 10²  m/s²  .

Andrei [34K]3 years ago
5 0

Answer:

it's b

Explanation:

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Given:

F = 3.0 x 10^-3 Newton

d = 6.0 x 10^2 meters

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Required:

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Formula:

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Solution:

So, to solve for Q2

 

Q2 = F • d²/ k • Q1

Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9 Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)

Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)

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Then, take the reciprocal of the denominator and start multiplying

Q2 = 1080 • 1 Coulombs/297

Q2 = 1080 Coulombs / 297

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