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3 years ago

If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?

Physics

2 answers:

sveta [45]3 years ago

5 0

The line that ought to be verified first in pushing the vessel

away from the dock in readiness to dock is the bow line. At the point when the bow line is

Further explanation:

verified, it is ideal to switch it and go to the dock, this will draw in the

line to fix such that will enable it to swing back in the dock.

Docking With Wind or Current from dock

Approach the dock gradually at a sharp edge (around 40 degrees).

Utilize turn around to stop when near the dock. Secure the bow line.

Put the pontoon in forward apparatus quickly, and gradually turn the guiding wheel hard away from the dock—this will swing in the stern. Secure the stern line.

Plane:

A point exists in zero measurements. A line exists in one measurement, and we indicate a line with two. A plane exists in two measurements. We determine a plane with three. Any two of the focuses determine a linea.

Thank you so much Hope will understand how this process will happen.

Answer Details:

Subject: Physics

Level: High School

Key Words:

Further explanation:

Docking With Wind or Current From the Dock:

Plane:

For further Evaluation

brainly.com/question/1156977

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Sphinxa [80]3 years ago

4 0

The line that should be secured first in pushing the boat away from the dock in preparation to dock is the bow line. When the bow line is secured, it is best to reverse it and turn to the dock, this will engage the line to tighten in a way that will help it swing back in the dock.

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Which one has greater potential energy? An 8 kg rock sitting on a 2.2 m cliff or a 6 kg rock sitting on a 3.2 m cliff.

Evgesh-ka [11]

Answer: The 6 kg rock sitting on a 3.2 m cliff.

Explanation:

The potential energy of an object of mass M that is at a height H above the ground us:

U = M*H*g

where g is the gravitational acceleration:

g = 9.8m/s^2

Then:

"An 8 kg rock sitting on a 2.2 m cliff"

M = 8kg

H = 2.2m

U = 8kg*2.2m*9.8 m/s^2 = 172.48 J

"a 6 kg rock sitting on a 3.2 m cliff"

M = 6kg

H = 3.2m

U = 6kg*3.2m*9.8m/s^2 = 188.16 J

You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.

6 0

2 years ago

A hole is drilled in a metal plate. When the metal is raised to a higher temperature, what happens to the diameter of the hole?

zysi [14]

Answer:

The diameter of the hole increases

Explanation:

Metals expand and contract with temperature. Whenever metal is heated, it usually expands in relation to its thermal expansivity. This expansion leads to a slight increase in surface area.

Once the surface area of the metal changes, this means that the dimensions of the whole metal surface changed. As a result, the diameter of the hole drilled in the metal plate will change also. In our case, the diameter of the hole will increase.

4 0

3 years ago

The orbital radius of an electron in a hydrogen atom is 0.846 nm. What is its de Broglie wavelength?

kotykmax [81]

Answer:

The value is $\lambda = 1.329 *10^{-9} \ m$

Explanation:

From the question we are told that

The orbital radius is $r = 0.846nm = 0.846 *10^{-9} \ m$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{2 * \pi r}{4}$

substituting values

$\lambda = \frac{ 2 * 3.142 * 0.846 *10^{-9}}{4}$

$\lambda = 1.329 *10^{-9} \ m$

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3 years ago

A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r

katen-ka-za [31]

Answer:

k = 167.33 N/m

Explanation:

The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
v = c = λ*f

where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =

4.92*10⁷ m.

Solving for f, we get the frequency of the radio waves:

f = 6.1 Hz

Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:

$\omega_{o}^{2} =\frac{k}{m} (1)$

Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

$\omega = 2*\pi *f (2)$

We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
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2 years ago

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Alik [6]

Answer:

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Explanation:

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3 years ago