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3 years ago

A sailboat is underway in the fog. what sound signal should you hear?

2 answers:

8 0

The conventional signal used by sailboats in conditions of reduced visibility such as heavy fog is one long blast followed by two short blasts.

The blasts help other boat operators locate one another's vessel in a condition where it is not easy to see. This signal is repeated in order to not only let others know of the vessel's position, but also help them know which way it is traveling. For example, if the blasts start to become distant, then the sailboat is travelling away from you.

Wittaler [7]3 years ago

4 0

The ordinary sign utilized by sailboats in states of reduced visibility for example heavy fog is one large Blasts followed by two short blast.

Further Explanation:

Horns in Foggy Conditions:

In foggy conditions you hear one delayed blast in addition to two short impacts at regular intervals of 2 minutes.

Personal Safety During Fog:

 Have all the team put on life jackets, as there will brief period if a collision happens

 Consider carefully before having the team cut on [safety harnesses], as they should probably to jump clear in a collision.

 On the off chance that the climate is unpleasant, the greater danger of falling overboard may require harnesses to be used.

 The life-raft should be at the ready for moment discharge if necessary.

 Have red and white flares nearby.

 In quiet climate, think about putting the dinghy over the side, [ towing it astern ] and being accessible in the case of a collision

Fog Forecasting:

An exact method for fog forecasting is the utilization of a hygrometer to discover the dew point temperature of the air.

Answer Details:

Subject: Physics

Level: High School

Key Words:

Horns in Foggy Conditions:

Personal Safety During Fog:

Fog Forecasting

For further Evaluation:

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A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

7 0

3 years ago

A train travels north at a speed of 50 m/s.

Gala2k [10]

The apparent velocity is B) 48 m/s north

Explanation:

Here we have a problem of relativity of velocities.

In fact, the train is travelling north at a speed of

$v_t = 50 m/s$

where this velocity is measured with respect to the ground.

At the same time, a passenger on the train is walking towards the rear (so, south) at a velocity of

$v'=2 m/s$

where this velocity is measured with respect to the train, which is in motion in the opposite direction.

Therefore, the apparent velocity of the passenger with respect to an observer standing on the ground is:

$v=V_t - v' = 50 - 2 = 48 m/s$

And the direction is north, since this number is positive.

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3 years ago

According to the Big Bang theory how long ago did the universe started??

Rashid [163]

According to the Big Bang theory how long ago did the universe started??
13 to 15 billion years ago

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3 years ago

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Conductors have ___<br> resistance.

vazorg [7]

Answer:

little/no

Explanation:

Conductors are materials, which conduct electricity and/or heat. That means, that their resistance to such energy is so little, that an electric current is able to pass through.

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3 years ago

Find the image position for a picture placed 3.0 cm outside the focal point of a converging lens with a 4.0 cm focal length. a.

horrorfan [7]

<span>Answer: Using 1/f = 1/d' + 1/d ...(where d' object distance and d is image distance) 1/4 = 1/7 + 1/d 1/4 - 1/7 = 1/d 3/28 = 1/d d = 28/3 d = 9.33 cm</span>

5 0

3 years ago