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goblinko [34]
3 years ago
9

two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing

of 130 degrees at 26 knots. How far apart are they after 1.5 hours?

Physics
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}

Put the value into the formula

v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}

v=\sqrt{18^2+26^2}

v=\sqrt{324+676}

v=10\sqrt{10}

We need to calculate the distance between the ships

d =v\times t

Put the value into the formula

d=10\sqrt{10}\times1.5\times1.852

d=87.84\ km

Hence, The distance between the ships is 87.84 km.

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
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Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

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The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

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a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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Answer:

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