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Angelina_Jolie [31]
4 years ago
14

Two ships leave a harbor at the same time. One ship travels on a bearing S 13 degrees W at 16 miles per hour. The other ship tra

vels on a bearing N 75 degrees E at 11 miles per hour. How far apart will the ships be after 2 ​hours?

Physics
1 answer:
Artemon [7]4 years ago
5 0

Answer:

47 miles

Explanation:

From the diagrammatic representation below;we can observe the illustration of the question.

after two hours;

The first ship = 16 miles per hour × 2 hours = 32 hours

the second ship = 11 miles per hour × 2 hours = 22 hours

to solve for ΔQPR in the diagram; we have;  

∠ P  = 13° + (2nd quadrant) + 15°

∠ P  = 13° + 90° + 15°

∠ P = 118°

To solve for the distance apart the two ship, we can see from the diagram that it is "p". So, using cosine rule; we have:

p² = q² + r² - 2qr Cos P

p² = 22² + 32² - 2(22)(32) Cos 118°

p² = 484 + 1024 - (1408) (-0.4695)

p² = 1508 + 661.056

p² = 2169.056

p = \sqrt{2169.056}

p = 46.5731252

p ≅ 47 miles

∴ The ships will be 47 miles  far apart after 2 hours.

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True Explosion in antarctic sea ice levels may cause another ice age.

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True

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The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

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Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

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i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

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