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larisa [96]
3 years ago
15

What should a sailboat operator do when approaching a PWC head-on

Physics
1 answer:
nekit [7.7K]3 years ago
8 0

<u><em>In accordance with the International Regulation for the prevention of collisions at sea</em></u><u>: </u>

<u>1.- A sailing boat has a passing preference over a motorized boat, </u><u>except when the motor boat is limited by its draft</u><u>. </u>

<u>2.- The sailboat must maintain its course and speed. </u>

<u>3.- </u><em><u>If it is evident that the PWC does not respond</u></em><u>, the sailboat must sound the warning signal, and change its course to starboard. </u>

<u>4.- </u><u><em>All actions must be taken as soon as possible</em></u><u>. </u>

<u>5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.</u>

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18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura  

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)  

P = (644.4 N, -613.075 N)

6 0
3 years ago
A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per
FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0
3 years ago
How does`ezzzzz321erfq23c2f
Nataliya [291]
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7 0
3 years ago
What is the name of the force that acts between any two objects because of<br>their masses?​
svetoff [14.1K]

Answer:

Explanation:

The force of attraction between 2 masses.

4 0
3 years ago
If you have a mechanical advantage of 2 with a block and tackle system.  How much force do you have to pull with?
Dovator [93]

Answer:

Half

Explanation:

You only have to exert a force equal to half the weight of the load to lift it.

7 0
3 years ago
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