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1 year 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]1 year 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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2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

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$\frac{1}{10}$ M

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To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

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$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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3 years ago

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a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

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And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

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