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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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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

7 0

3 years ago

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

2 years ago

The period of a 261-Hertz sound wave is

SCORPION-xisa [38]

The period of a 261 hertz sound wave is
.000383 seconds

7 0

3 years ago

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Assoli18 [71]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{70}{7} = 10 \\$

We have the final answer as

Hope this helps you

4 0

3 years ago

In world war 1 , Multiple reflection was used in submarines. How multiple reflection works and helps in such situation?

prisoha [69]

<u>Answer:</u>

Submarines use a device called Periscope that uses the concept of multiple reflections and help us see objects above the water surface.

<u>Explanation:</u>

When a light ray falls on a reflecting surface like a mirror, it gets reflected. In multiple reflections, the incident light ray is made to reflect multiple times by arranging the reflecting surfaces in different ways.

In submarines, we use Periscope, which is a long tube like structure. The long tube is bent at ends. It uses two simple mirrors which are placed parallel to each other at an angle of 45 degrees. The light from one mirror gets reflected to the other mirror, thus causing a multiple reflection.

8 0

3 years ago