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dimaraw [331]
4 years ago
8

Ep-40 if you run aground in an outboard boat and you are not taking on water, what is the first step in attempting to free your

vessel?
Physics
1 answer:
ehidna [41]4 years ago
3 0

In order to free your vessel when you run aground in an outboard boat and you are not taking on the water, the very first thing you need to do is just to stop the boat’s engine to stop it from moving then try to lift the out drive, and shift the weight away from the impact zone.  Once this is done, you can already now assist the situation and you will be sure that the bat will no longer take into the water.  You can now set an anchor in order for you to be safe from the force that can push you and the boat further.  You can then use the boat hook so you can check how deep the water is around you.

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Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s
vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

\Delta y = \frac{x\lambda}{d}

Where

\Delta y = Distance between fringes

x = distance between slits and screen

d = Distance between slits

\lambda= Wavelength

Our values are given as

d= 0.08mm

x =3m

\lambda = 633nm

In this way replacing in the equation,

\Delta y = \frac{x\lambda}{d}

\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}

\Delta y = 2.37*10^{-2}m

\Delta y = 2.37cm

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

\Delta Y_2 = \frac{\Delta Y_1}{n}

\Delta Y_2 = \frac{2.37}{1.33}

\Delta Y_2 = 1.78cm

3 0
4 years ago
PLEAASE HELP KE WITH THESE THREE YOULL GET POINTS
lozanna [386]

Answer:

  1. 0.0121
  2. 19.8387
  3. 2.2632

Explanation:

1.

\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{0.01223}{1.01}\:\\\mathrm{is}\:0.0121

2.

\frac{\begin{matrix}\space\space&\textbf{\space\space}&\space\space&\space\space&\space\space&4&9&10\\ \space\space&\textbf{1}&9&.&8&\linethrough{5}&\linethrough{10}&\linethrough{0}\\ -&\textbf{0}&0&.&0&1&1&3\end{matrix}}{\begin{matrix}\space\space&\textbf{1}&9&.&8&3&8&7\end{matrix}}\\\\=19.8387

3.

0.1886\times 12\\\\Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer\\1886\times\:12=22632\\\\0.1886\mathrm{\:has\:}4\mathrm{\:decimal\:places}\\12\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}4\mathrm{\:decimal\:places}\\\\=2.2632

3 0
3 years ago
Can you put this image back together? Type the correct order of letters below.
Art [367]

Answer:

CABD

.....................

6 0
2 years ago
Technician A says a special puller and installer tool is required to remove and install the vibration damper. Technician B says
andrey2020 [161]

Answer:

Both of them are correct

Explanation:

Here both the technician are correct. A special puller and installer tool is required to remove and install the vibration damper. Removing and installing a damper is not an easy task, it requires a definite set of technique and tools.

According to Technician B if the inertia ring on the vibration damper is loose, the damper must be replaced is absolutely correct, inertia ring once loosened cannot be tightened. hence, we have change the the damper.

6 0
3 years ago
Which force requires contact?
GuDViN [60]

Answer:

I think its D too

Explanation:

3 0
4 years ago
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