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3 years ago

What is most important when selecting the right pfd for a passenger on board your boat?

2 answers:

3 0

The most essential while choosing the privilege PFD for a traveler is the information required for choosing right PFD for the traveler's body weight and trunk size.Note that the offspring of what age under 8 years old are required to wear legitimately secured PFD while on vessel unless in securely closed cabin.

Nostrana [21]3 years ago

3 0

Answer:

Passenger's body weight and chest size

Explanation:

Finding a PFD (Personal Flotation Device) for a passenger requires you to find a PFD which fits the person properly for his/her required weight. There are three categories of sizes:

Men's size varies from small to 3x large where the chest size varies and the weight of the person is over 90 lbs or 41 kg

Ladies size varies from x small to x large where the chest size varies and the weight of the person is over 90 lbs or 41 kg

Kids weight varies from 90 lbs or 41 kg to 30 lbs or 14 kg

The PFD need to snugly fit to the person's body. So, to get the best fit you need to find the weight of the person and the corresponding chest size.

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Determine the velocity (in m/s) of the object during the first three seconds. Include a numerical answer accurate to the first d

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If it’s moved 3m in 3s then it should be 3m/s

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3 years ago

Where is the energy in a glucose molecule stored?

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Answer:

Energy is stored in the bonds between atoms

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3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

the answer to (b) is 6.86m

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

the answer to (c) is 2.86s

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3 years ago

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Answer:

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It is not a statement or direct application of the second law of thermodynamics. This statement is in accordance with first law of thermodynamics.

All the other two statements are in accordance with second law of thermodynamics.

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3 years ago

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8090 [49]

Transversal wave is the answer

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3 years ago

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