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Dimas [21]
3 years ago
11

A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity rela

tive to the Earth is 4.80 m/s 5.00º west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)
Physics
1 answer:
ASHA 777 [7]3 years ago
8 0

Answer:

Velocity of Gulf Stream= 34.5^\circ west of north at a speed of 2.03\ \rm m/s

Explanation:

Given

  • speed of ship relative to Gulf stream=25^\circ  west of north at a speed of 4 m/s
  • speed of ship relative to earth=5^\circ  west of north at a speed of 4 .8 m/s

Let V_S be the velocity of the ship relative to the earth and V_{sg} be the velocity of the ship with respect to the Gulf stream

Let the west direction be  negative x axis and north direction be positive x axis

Now

V_{sg}=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\V_s-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\-4.8\sin5^\circ \vec i+4.8\cos5^\circ \vec j-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\Vg=-1.68\vec\ i+1.156\vec\ j\ \rm m/s\\

magnitude of the velocity is given by

V_g=\sqrt{(1.68^2+1.156^2)}\\V_g=2.03\ \rm m/s\\

Let the velocity of gulf stream makes an angle \theta with the positive y axis we have

\tan\theta=\dfrac{1.156}{1.68}\\\theta=34.5^\circ

Velocity of Gulf Stream 34.5^\circ west of north at a speed of 2.03\ \rm m/s

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
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Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

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Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

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For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

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For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

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