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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:

brainly.com/question/5144573

brainly.com/question/112785

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A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

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3 years ago

An industry uses natural gas for manufacturing and uses the waste heat to produce electricity. This is an example of which proce

Lana71 [14]

answer is chemical process

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3 years ago

what will happen to the brightness of the other bulbs in a series circuit if any of the bulbs were to burn out?

Ivenika [448]

Answer:

not work.

Explanation:

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3 years ago

Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

$a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N$

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

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2 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$