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mixer [17]
2 years ago
10

A car that is traveling in a straight line at 40 km/h can brake to a stop within 20 m. If the same car is traveling at 120 km/h

what would be its stopping distance in this case? Assume the braking force is the same in both cases and ignore air resistance.
Physics
1 answer:
stiks02 [169]2 years ago
8 0

Answer:

180 m

Explanation:

Case 1.

U = 40 km/h = 11.1 m/s, V = 0, s = 20 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0 = 11.1 × 11.1 - 2 × a × 20

a = 3.08 m/s^2

Case 2.

U = 220 km/h = 33.3 m/s, V = 0

a = 3.08 m/s^2

Let the stopping distance be x.

Again use third equation of motion

0 = 33.3 × 33.3 - 2 × 3.08 × x

X = 180 m

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0
3 years ago
A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
3 years ago
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