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3 years ago

How do noise and vibration affect you when operating a boat?

Physics

2 answers:

natta225 [31]3 years ago

4 0

Answer:

It will slow down your reaction time, and reducing ability to control the boat and it will give physical fatigue .

Explanation:

When you aboard a boat you are subjected to both low and high frequencies from the wind and from the engine of the boat. You can feel the waves which are of low frequency as a vibration.

Like other machines boat can produce noise and vibration together that will make you tired, and in some cases your reaction times will slow down and you will loose the ability to control that boat.

After that the heat and glare from the sun will also make you physically fatigue and in this condition operating a boat is more dangerous.

kogti [31]3 years ago

3 0

The noise and vibration could affect your body and mood in many ways while operating a boat. The noise and vibration could make you moody or uncomfortable, because it will also distract you in doing what you're supposed to do. Your body will also be affected, making your body tired because of the things surrounding you and the noise and vibration, it could affect your body in many ways.

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A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

3 years ago

Determine the magnitude of the component of F directed along the axis of AB. Set F = 520 N .

Soloha48 [4]

Answer:

The component of F along AB is equal to Fcos45

F = 520N

Component along AB = 520cos45

= 367.7N

This is done by rotating the diamonds such that AB is now taken as the x-axis. Then the force F is resolved along AB.

Explanation:

7 0

3 years ago

Read 2 more answers

Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong):

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong) we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) Asumming (a + b)·b instead a+b·b we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

5 0

4 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$