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2 years ago

You are operating a powerboat at night. Your red sidelight must be visible to boats approaching from which direction(s)

Physics

1 answer:

Dmitriy789 [7]2 years ago

4 0

The answer is the red sidelight on a powerboat should be visible from the front and from the left (port side).

What are Sidelights?

There is various combinations of lights that must be used on a boat when it is dark, and these are:
Sidelights: These lights are called combination lights and are red and green. The red sidelight must be visible from the port side and the green light indicates the right side (the starboard).
Stern light: The stern light is seen at the back end of the vessel.
Masthead Light: The masthead light is a white light that shines forwards and on all sides of the vessel. All powered vessels must use this light.
All-Round white light: This light is the major light that is used to join the masthead light and the stern light. This single light is visible to all vessels from all directions.
Thus, as a rule for a boat rider, he should show the vision of red light and it should be visible from the front and from the left (port side).

To learn more about Sidelights visit:

brainly.com/question/28205057

#SPJ4

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A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway

Andru [333]

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

5 0

3 years ago

What should he do next in order to follow the steps of inquiry?

andrezito [222]

Since he wonders, then according to the formal, methodical procedures of scientific inquiry, he should not go running off and trying stuff before he forms a hypothesis that can be tested.

In his mind, at some point, he should say to himself "Self ! It seems to me that if salt is added to water, it makes the water boil at a higher temperature than pure water does."

And only THEN, with this statement in mind, he's ready to design an experiment to test it.

6 0

3 years ago

chapter 2 linear motion problems a student launches an arrow upward with an unknown initial velocity. the arrow takes 2.3 second

anzhelika [568]

Answer:

$V_{0}= 22.5\frac{m}{s}$ and Ymax=25.8m

Explanation:

Velocity at any time is given by $V=V_{0}sin\theta -gt$ . but when the arrow is on the top its velocity is zero and if it is launched upward the angle is 90°, so.

$0=V_{0} sin90-gt$

$V_{0}=gt=9.8\frac{m}{s^{2}}.2.3s=22.5\frac{m}{s}$

At the maximun height, position is given by $Ymax=V_{0}sin\theta.t-\frac{1}{2}gt^{2}$ , replacing $Ymax=22.5\frac{m}{s}x2.3s-\frac{1}{2}x9.8x(2.3)^{2}=25.8\frac{m}{s}$