In which of these situations would there be the least kinetic energy?

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

Dmitriy789 [7]

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:

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4 0

1 year ago

HELP PLS 100 POINTS TAKING TEST RN

s2008m [1.1K]

Answer:

I think he answer is C but I could be wrong

Explanation:

brainliest plz

8 0

3 years ago

Read 2 more answers

Which of the following items has the most inertia while at rest?

netineya [11]

Answer:

airplane

Explanation:

as greater mass greater inertia

8 0

3 years ago

Which of the following represents the wavelength of a wave?

dlinn [17]

Answer: D. the distance between the highest points of consecutive waves

Explanation:

The wavelength of a wave is defined as the <em>distance traveled by a periodic perturbation that propagates through a medium in a given time interval</em>. It is usually represented by $\lambda$ and can be calculated if the frequency of the wave is known, since there is an inverse relationship between both.

In the specific case of a periodic sine wave (which is the way in which a wave is usually represented graphically) the wavelength can be determined as the distance between two consecutive maxima of the disturbance.

Therefore, the correct option is D.

8 0

3 years ago

Read 2 more answers

A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the

Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$ ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$ (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

7 0

3 years ago