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zysi [14]
3 years ago
9

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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Explanation:

Length= 2cm= 20mm

Now meter stick can read to nearest millimeter.

It is given that length is to be measured with a precision of 1% of 20mm= 1/100 * 20= 0.2mm

Since the least count is 1mm of meter stick and precision required is less than that. So, meter stick cannot be used for this, travelling microscope can be used for this as it can read to 0.1mm.

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A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2
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(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

  • Z = √[R²+(2πfL)²]........................ Equation 1

Where:

  • R = resistance of the coil
  • f = Frequency
  • L = Inductance of the coil
  • Z = Overall impedance of the circuit

From the question,

Given:

  • R = 7 ohms
  • L = 31.8 mH = 0.0318 H
  • f = 50 Hz
  • π = 3.14

Substitute these values into equation 1

  • Z = √[7²+(2×3.14×50×0.0318)²]
  • Z = √(49+99.7)
  • Z = √(148.7)
  • Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

  • I = V/Z.................. Equation 2

Where:

  • V = Voltage
  • I = current in the circuit.

Given:

  • V = 230 V.

Substitute into equation 2

  • I = 230/12.19
  • I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

  • ∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

  • ∅ = Phase angle.


Substitute into equation 3

  • ∅ = tan⁻¹(2×3.14×50×0.0318/7)
  • ∅ = tan⁻¹(9.9852/7)
  • ∅ = tan⁻¹(1.426)
  • ∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

  • pf = cos∅............ Equation 4

Where:

  • pf = power factor.

Substitute the value of ∅ into equation 4

  • pf = cos(54.97°)
  • pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

  • P = V×I×pf................ Equation 5

Where:

  • P = The power consumed

Substitute the values into equation 5

  • P = 230(18.87)(0.574)
  • P = 2491.22 W


Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

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