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rewona [7]
4 years ago
8

What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your

answer to two significant figures and include the appropriate units?
Physics
1 answer:
torisob [31]4 years ago
6 0
<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R 
I = MR² = (0.200)(5) = 1.00 kg-m² 
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal 
V = 3.0 m/s 
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s 
w = [V component / R] = 1.3433/√5 = 0.601 rad/s 
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s</span><span>


i hope I'm right</span>
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svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

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Explanation:

Given:

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<u>Now the horizontal component of velocity:</u>

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<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

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v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

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Read 2 more answers
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