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vichka [17]
3 years ago
11

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

Physics
2 answers:
Rufina [12.5K]3 years ago
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:

brainly.com/question/5144573

brainly.com/question/112785

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A ball is thrown horizontally from the top of a building at 2 m/s. It takes 3 seconds to reach the ground. How far did the ball
svlad2 [7]
Every second it travels 2 meters and it traveled 3 so (2 x 3) would be 6meters it traveled
4 0
3 years ago
A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m
Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

I_{disk} = \frac{1}{2} mr^2\\\\I_{disk}  = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk}  = 0.305 \ kgm^2

The moment of inertia of the rod about its center is calculated as follows;

I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2

The initial rotational kinetic energy of the disk and rod;

K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i=  \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \  \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J

The final rotational kinetic energy of the disk-rod system is calculated as follows;

K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J

The loss in rotational kinetic energy due to the collision is calculated as follows;

\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J  \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

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c adding research resources during an investigation

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Nezavi [6.7K]
We have F = kx or ma = kx where m and k are constants. Therefore, if x is halved, a must be halved too.
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Answer:A

Explanation:

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