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4 years ago

A boat is leaving the dock. what sound signal should be used to alert others?

2 answers:

8 0

Blasts are used for communicating different signals when vessels move.When a boat is leaving, a long blast, at intervals of not more than two minutes, is used for alerting other vessels, when a boat is sailing. When leaving the dock, only a single prolonged blast is used for indicating others, that you are leaving.

crimeas [40]4 years ago

6 0

One longed blast is a notice signal. Five short, fast blasts are utilized signal danger or to signal that you don't comprehend or you can't help with the other boater’s intentions.

Further Explanations:

Sound sign:

Let different boaters know where you are situated during times of restricted visibility, such as extreme fog. If you hear the mist sign of a vessel you can't see, slow to a base speed until you are sure certain there isn't a danger of collision.

One delayed blast less than 2 minutes:

One delayed blast at intervals of not over two minutes is the sign utilized by power-driven vessels when underway.

One delayed blast plus 2 Short blasts:

One delayed blast in addition to two short blasts at intervals of not over two minutes is the sign utilized by sailing vessels.

One delayed Blast:

One delayed blast is a notice signal (for instance, utilized when coming around a blind bend or leaving the dock.

At least five short blasts:

(At least five) short, fast blasts are utilized to signal danger or to signal that you do not understand or you can't help with the other boater’s aims.

Answer Details:

Subject: Physics

Level: High School

Key Words:

Sound sign

One delayed blast less than 2 minutes

One delayed blast plus 2 Short blasts

One delayed Blast

At least five short blasts

For further Evaluation:

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For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

If a 15 N box is lifted a distance of 3 m, how much work is done?

Naily [24]

Answer:

W=45J

Explanation:

W=Fd

W=15(3)=45

W=45J

7 0

3 years ago

Read 2 more answers

A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so

hodyreva [135]

Answer:A) Risk(R)= $1000

B) There is justification for spending an additional cost of $100 to prevent a corrosion whose consequence in monetary terms is $1000

Explanation:R= Risk,

P=Probability of failure

C= Consequence of failure

Mathematically, R=P ×C

10 out of 1000 carbon-steal products failed

Probability of failure= 10/1000 =0.01

The consequence of failure by corrosion given in monetary term =$100,000

Risk of failure = 0.01 × $100,000

R=$1000

4 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

$d=v_{0}.t + \frac{1}{2} .a.t^{2}$ , where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + $\frac{1}{2}.10.t^{2}$

h = 5t²

$t_{T}$ = $\sqrt{\frac{h}{5} }$

On the other planet:

h = 0.t + $\frac{1}{2}.30.t^{2}$

h = 15.t²

$t_{P}$ = $\sqrt{\frac{h}{15} }$

Comparing the 2 planets:

$\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }$

$\frac{t_{T} }{t_{P} }$ = $\sqrt{3}$ or $t_{T} = \sqrt{3}.t_{P}$

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

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3 years ago

A drawback of burning biomass to produce electricity is that it is ___.

attashe74 [19]

Answer: Unsustainable wood harvesting can lead to deforestation, soil erosion, and desertification.

Explanation: you welcome

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3 years ago