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Pavlova-9 [17]
3 years ago
15

Two male moose charge at each other with the same speed and meet on a icy patch of tundra. As they collide, their antlers lock t

ogether and they slide together with one-third of their original speed.
What is the ratio of their inertias?
Physics
2 answers:
Digiron [165]3 years ago
7 0

change in velocity of larger moose: (1/3)v - v = -(2/3)v

change in velocity of small moose: (1/3)v - (-v) = (4/3)v

(change in velocity of larger moose)/(change in velocity of smaller moose) = 2


Explanation:

It is the "ma" in F=ma that we tend to call inertia.Inertia is that the resistance of any entity to any amendment in its position and state of motion.

his includes changes to the object's speed, direction, or state of rest. Inertia is additionally outlined because the tendency of objects to stay getting a line at a continuing speed.

Ksenya-84 [330]3 years ago
4 0
Change in velocity of larger moose: (1/3)v - v = -(2/3)v 
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.



</span>
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a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

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Explanation:

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  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

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