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frozen [14]
3 years ago
11

EJ is taking his little sister, Ebony, on her first carousel ride. His mass is 75.0 kg, and her mass is 10.0 kg. They ride the s

ame horse, and it is 11.0 meters from the center of the carousel. If their angular momentum is 14020 kg m2/s, what is their linear speed?
Physics
1 answer:
Alika [10]3 years ago
6 0

In this question we have given

mass of Ebony=75Kg

His sister mass=10kg

here total mass=mass of Ebony+his sister mass

m=75+10  

m=85kg

radius of circular path=11m

angular momentum=14020kgm^2/s

we know that,

angular momentum=interia*angular speed

angular momentum=I*w...........(1)

here, interia  I=mr^2  

and w=v/r (here v is the linear velocity and r is radius)

put  values of I and w in equation (1)

angular momentum=mr^2*v/r

angular momentum = mvr  

14020 = (75+10) x v x 11  

v = 14020/(85x11) = 14.99m/s







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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

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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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