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ICE Princess25 [194]
3 years ago
11

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

ips by calculating their velocity just after they collided. According to the last radio transmission from the 40,000-ton luxury liner, the Hedonist, it was going due west at a speed of 20 knots in calm seas through a rare fog just before it was struck broadside by the 60,000-ton freighter, the Ironhorse, which was traveling directly northeast at 10 knots. The transmission also noted that when the freighter's bow pierced the hull of the liner, the two ships stuck together and sank.To determine the search area, find the final velocity vector just after moment of collision.
Physics
1 answer:
Leya [2.2K]3 years ago
6 0

<u>Solution and Explanation:</u>

The following calculation is made in order to find out the area and the final velocity vector.

Using the given information and the data in the question,  

m1u1 + m2u2 = (m1 + m2)  multiply with v

40000 multiply with ( -20i ) + 60000 multiply with ( 10j ) = 100000 multiply with v

  Therefore, v = -8i + 6j

That is  |v| = 10 knots towards the 36.86 degree north of the west

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Answer:

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Explanation:

Warmer air is less dense than cold, which is why warm air tends to rise and cold air sinks. Being acted on by gravity, colder, denser air weighs more and exerts greater pressure per unit area.

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At the point of fission, a nucleus of ^235 U that has 92 protons is divided into two smaller spheres, each of which has 46 proto
IrinaK [193]

Answer : F = 3.5\times10^{3}\ N

Explanation :

Given that

Radius of sphere r = 5.90\times 10^{-15}\ m

The distance between the centers of the two spheres is

r = 2\times 5.90\times 10^{-15}\ m

The charge of the sphere q = 46\times1.6\times10^{-19} C

The magnitude of the repulsive force between the charges pushing them a part is

Using coulomb law

F = \dfrac {kq_{1}q_{2}}{r^{2}}

F = \dfrac{9\times10^{9}\times (46\times1.6\times10^{-19})^{2}C^{2}}{2\times(5.90\times10^{-15})^{2}\ m^{2}}

F = 3501.3\ N

F = 3.5\times10^{3}\ N

Hence, this is the required solution.









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3 years ago
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The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t&gt;0 , where is measured in seconds and
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The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at 0.

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>
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The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s

The velocity is positive at when the time of motion is as follows;

0.

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2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m

Learn more about motion of particles here: brainly.com/question/11066673

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