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3 years ago

Why is it important to always be wearing a personal flotation device (PFD) when in or around the water?

Engineering

2 answers:

oee [108]3 years ago

4 0

The correct answer is C. A PFD takes some time to fit properly especially in an emergency.

Explanation:

The purpose of a Personal Flotation Device (PFD) such as a life jacket is to keep people floating to prevent drowning. Additionally, its effectiveness depends on whether the Personal Floating Device was fitted properly. Because of this, if an activity involves being in or around water, it is important to wear a Personal Flotation Device and to fit this properly before approaching the water because it is difficult to adjust the device during an emergency. Thus, it is important to wear a PFD in or around water because this takes some time to fit properly (Option C.)

IgorLugansk [536]3 years ago

3 0

Answer:Wearing a PFD reduces the chances of capsizing your vessel

Explanation: I THINK

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Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i

DedPeter [7]

Answer:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

Explanation:

The internal energy is equal to:

$dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV$

The entropy is equal to:

$dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV$

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

$\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }$

Replacing:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

4 0

3 years ago

A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of

vlabodo [156]

Answer:

$S_{gen} = 18.519\,\frac{kJ}{K}$

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

$W_{heater} + U_{sys,1} - U_{sys,2} = 0$

$W_{heater} = U_{sys,2} - U_{sys,1}$

$W_{heater} = m\cdot (u_{2}-u_{1})$

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

$S_{sys,1} - S_{sys,2} + S_{gen} = 0$

$S_{gen} = S_{sys,2} - S_{sys,1}$

$S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the steam are obtained from steam tables:

Intial State

$P = 200\,kPa$

$T = 120.21\,^{\textdegree}C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 1010.7\,\frac{kJ}{kg}$

$s = 2.9294\,\frac{kJ}{kg\cdot K}$

$x = 0.25$

Final State

$P = 869.567\,kPa$

$T = 173.88\,^{\textdegree} C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 2578.6\,\frac{kJ}{kg}$

$s = 6.6332\,\frac{kJ}{kg\cdot K}$

$x = 1.00$

The entropy change of the steam during the process is:

$S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)$

$S_{gen} = 18.519\,\frac{kJ}{K}$

8 0

3 years ago

Para uma dada velocidade do vento, o gerador de turbina mostrado na figura produz 22 kW de potência elétrica. Como um sistema co

____ [38]

Answer:

Explanation:

4 0

3 years ago

‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?

Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

$E=12\times 9.8\times 25\\\\E=2940\ J$

So, the potential energy of the mass is 2940 J.

3 0

3 years ago

Maslow’s highest need.

soldi70 [24.7K]

Answer:

Self-actualization

Explanation:

8 0

3 years ago