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vitfil [10]
3 years ago
5

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

h mechanical energy was dissipated as this happened? The coefficient of kinetic friction between the tires and the road is 400.40un.
Physics
1 answer:
Harman [31]3 years ago
3 0

Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

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A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /
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The following three questions refer to a situation in which a driver is in a car that crashes into a solid wall. The car comes t
Andreas93 [3]

Answer:

1) p₀ = 45000 N / s ,   p₀ '= 1800 , b)  I = -45000 N s ,  I = 1800 Ns

Explanation:

Impulse equals the change in momentum

         I = Δp

1) the initial moment of the car

         p₀ = M v

          p₀ = 1500 30

          p₀ = 45000 N / s

the change at the moment is

          Δp = 45000

because the end the car is stopped

moment of the person

          P₀ ’= m v

         p₀ '= 60 30

          p₀ '= 1800

          D₀ '= 1800

2) of the momentum change impulse ratio

        car

              I =  Δp

              I = -45000 N s

        person

              I = Δpo '

              I = 1800 Ns

3) the object that give the momentum to stop the wall motoring

The person is stopped by the impulse given by the car

a) This area is the one that absorbs most of the vehicle impulse

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A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14
Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

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(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

v_{B} = \sqrt{237.16}

vB = 15.4 m/s : speed of the cart at B

4 0
3 years ago
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