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dedylja [7]
4 years ago
9

una bala de 20 g choca con un fango como se muestra en la figura y penetra una distancia de 6 cm antes de detenerse. calcule la

fuerza de frenado f si la velocidad de entrada fue de 80m/s​
Physics
1 answer:
aleksklad [387]4 years ago
7 0

Answer:

A 20g bullet collides with a mud as shown in the figure and penetrates a distance of 6cm before stopping. calculate the braking force f if the input speed was 80m/s

Explanation:

Given that,

A bullet of mass

M = 20g = 0.02kg

The bullet is fired into a mud initially at rest

The bullet penetrate a distance of 6cm in the mud

S = 6cm = 0.06m

The input velocity into the mud is 80m/s, this implies that, the initial velocity is 80m/s

U = 80m/s

So, the bullet stop after a distance of 6cm, so the final velocity is 0m/s

V = 0m/s

So, we need to find the braking force

Breaking force (F) is given as

F = ma

Where m is mass and 'a' is deceleration

So, we need to find the deceleration, using equation of motion

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m/s²

So, the negative sign shows that the bullet is decelerating

So, a = 53,333.33 m/s²

So, breaking force

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

The breaking force is 1066.67 N

In spanish

Dado que,

Una bala de masa

M = 20 g = 0.02 kg

La bala se dispara en un lodo inicialmente en reposo

La bala penetra una distancia de 6 cm en el barro.

S = 6 cm = 0.06 m

La velocidad de entrada al lodo es de 80 m / s, esto implica que la velocidad inicial es de 80 m / s

U = 80 m / s

Entonces, la bala se detiene después de una distancia de 6 cm, por lo que la velocidad final es de 0 m / s

V = 0 m / s

Entonces, necesitamos encontrar la fuerza de frenado

La fuerza de ruptura (F) se da como

F = ma

Donde m es masa y 'a' es desaceleración

Entonces, necesitamos encontrar la desaceleración, usando la ecuación de movimiento

V² = U² + 2as

0² = 80² + 2 × a × 0.06

0 = 6400 + 0.12a

0.12a = -6400

a = -6400 / 0.12

a = - 53,333.33 m / s²

Entonces, el signo negativo muestra que la bala se está desacelerando

Entonces, a = 53,333.33 m / s²

Entonces, fuerza de ruptura

F = ma

F = 0.02 × 53,333.33

F = 1066.67 N

La fuerza de ruptura es 1066.67 N

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Answer:

The coefficient of friction is (F/(19.6·m)

Explanation:

The given parameters are;

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Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 46.0 mph . Beth leaves Los A
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