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Scilla [17]
4 years ago
14

On dry concrete, a car can decelerate at a rate of 7.00 m/s2 , whereas on wet concrete it can decelerate at only 5.00 m/s2 . Fin

d the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.
Physics
1 answer:
MariettaO [177]4 years ago
3 0

Answer:

Explanation:

a ) Let the distance required in former case be d₁ .

initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 7 x d₁

d₁ = 64.28 m

b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 5 x d₂

d₂ = 90  m

c)

t = .5 s

s₁  = ut - .5 at²

= 30 x .5 - .5 x 7 x .5²

= 15 - .875

= 14.125 m

t = .5 s

s₂  = ut - .5 at²

= 30 x .5 - .5 x 5 x .5²

= 15 - .625

= 14.375  m

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La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s

Explanation:

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El coeficiente de fricción, μ = 0,15

La altura a la que debe ascender el trineo, h = 4 m

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g = Th aceleración debida a la gravedad ≈ 9.81 m/s²

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El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules

El trabajo total requerido para levantar el trineo 4 metros por la rampa, W_t = W_f + P.E.

W_t = 20.4 J + 78.84 J ≈ 99.24 J

Energía cinética, K.E. = 1/2 × m × v²

La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;

K.E. = 1/2 × 2 kg × v² ≈ 99.24 J

∴ v² ≈ 99.24 J/(1/2 × 2 kg)

La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s

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