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3 years ago

How should you approach a dock when the wind or current is pushing you away from the dock?

Physics

1 answer:

erastova [34]3 years ago

8 0

Answer:

If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.

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2 years ago

Read 2 more answers

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

Soloha48 [4]

Answer:

$\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}$

$\boxed{\sf Distance \ covered \ (s) = 100 \ m}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

$\sf (i) \ From \ 1^{st} \ equation \ of \ motion:$

$\sf \implies v = u + at$

$\sf \implies 20 = 0 + a(10)$

$\sf \implies 10a = 20$

$\sf \implies \frac{10a}{10} = \frac{20}{10}$

$\sf \implies a = 2 \: m/ {s}^{2}$

$\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:$

$\sf \implies s = ut + \frac{1}{2} a {t}^{2}$

$\sf \implies s = (0)(10) + \frac{1}{2} \times 2 \times {(10)}^{2}$

$\sf \implies s = \frac{1}{ \cancel{2}} \times \cancel{2} \times {(10)}^{2}$

$\sf \implies s = {10}^{2}$

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3 years ago

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Shtirlitz [24]

Answer:

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2 years ago