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snow_lady [41]
3 years ago
13

Two boats are operating in the same general area. who is responsible for avoiding a collision?

Physics
2 answers:
Vaselesa [24]3 years ago
7 0

The operators of both boats are responsible for collision. Individual skippering a boat must do everything possible to avoid collision.

Part of the navigation rule states that every person skippering a boat should consider all dangers and collision risks and that could also mean breaking the rules if evasive action is required.

<h2>Further Explanation</h2>

There are certain rules that every boat operator must follow when confronting other vessels.  These rules are very essential so as to avoid collision.

The navigation rules are below

  • If two boats are getting close to each other head on, it is important for both boats to change their course to the right (starboard) in order for both boats to end up passing port-side to port-side
  • In a situation that two boats cross each other part and are on course for collision, it must important to give way to the boat on the right (starboard). The boat giving way must take prompt action. Such boat operator can either stop the boat or change the course to the right.
  • In the circumstance that the boat giving way has another boat on the port hand side that is not changing course or give way, the helmsman can take action with a stop or change direction
  • It is important that any boat, motorized, sail or powered must give plenty of room if overtaking.

LEARN MORE:

  • avoid a collision between two boats brainly.com/question/4170661

KEYWORDS:

  • boats
  • vessels
  • navigation rules
  • collision
  • maintain course
goblinko [34]3 years ago
6 0
In this case, the one that responsible for avoiding a collision would be: <span>the operators of both boats
When the operators of each boat spot each other in a same area, they should use their siren to notify each other's position and uses communication device to determine how they should pass through to avoid collision</span>
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The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

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5 0
3 years ago
Can someone help? Please?
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Answer:

A. Speed

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Read 2 more answers
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
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