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Natali5045456 [20]
3 years ago
11

A car is traveling at 42.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0

.107, what is the minimum distance in which the car will stop?
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

64.85 m

Explanation:

a)

μ = Coefficient of friction between road and tire on rainy day = 0.107

g = acceleration due to gravity = 9.8 m/s²

a = acceleration experienced by car due to friction = - μg = - (0.107) (9.8) = - 1.05 m/s²

v₀ = initial velocity of the car = 42 km/h = 11.67 m/s

v = final speed of the car = 0 m/s

d = minimum distance traveled before stopping

Using the equation

v² = v₀² + 2 a d

0² = 11.67² + 2 (- 1.05) d

d = 64.85 m

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Tenemos.

Masa de la mesa = 20kg
Masa encima = 10kg
Masa total = 30kg
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Presió(P)n que se ejerce sobre cada pata.

P =F/A
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P = 294kg * m/s²/0,002m² Pero kg *m/s² = Nw
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Respuesta.
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