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mihalych1998 [28]
3 years ago
9

Un auto de 1000 kg que lleva una velocidad de 72 km/h se detiene en una distancia de 100 metros cuando se le aplican los frenos.

(a) ¿Cuál es la fuerza constante producida por los fre. Nos? (b) Con esta fuerza, ¿cuál seria la distancia de frena do si la velocidad hubiera sido de 36 km/h?
Physics
1 answer:
Flura [38]3 years ago
6 0

Answer:

a)    F = 2000 N , B)   x = 25 m

Explanation:

a) To solve this exercise we can use the relationship between work and kinetic energy

          W = ΔK

The job is

           W = -F x

the negative sign is because the force of the brakes is contrary to the movement

as the car stops its final kinetic energy is zero

           K = ½ m v²

let's substitute

         - F x = 0 - ½ m v²

            F = ½ m v² / x

Let's reduce the magnitudes to the SI system

           v = 72 km / h (1000m / 1km) (1 h / 3600 s) = 20 m / s

let's calculate

          F = ½ 1000 20²/100

          F = 2000 N

b) x = ½ mv2 / F

let's slow down to the SI system

         v = 36 km / h = 10 m / s

let's calculate

             x = ½ 1000 10²/2000

             x = 25 m

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
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Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

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-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

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3 years ago
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