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crimeas [40]
3 years ago
5

A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s

olid disc with a radius of 2.50 m and a mass of 10.0 kg is spinning at 3500. rpm. The smaller disc is carefully pressed against the larger disc (flat side to flat side) so that they spin together without slipping. What is the angular velocity of the large disc
Physics
1 answer:
drek231 [11]3 years ago
4 0

Answer:

Explanation:

This problem is based on conservation of angular momentum.

moment of inertia of larger  disc I₁ = 1/2 m r²  , m is mass and r is radius of disc . I

I₁ = .5 x 20 x 5²

= 250 kgm²

moment of inertia of smaller  disc I₂ = 1/2 m r²  , m is mass and r is radius of disc . I

I₂ = .5 x 10 x 2.5²

= 31.25 kgm²

3500 rmp = 3500 / 60 rps

n = 58.33 rps

angular velocity of smaller disc ω₂ = 2πn

= 2π x 58.33

= 366.3124 rad /s

applying conservation of angular momentum

I₂ω₂  = ( I₁ +I₂) ω  , ω is the common angular velocity

31.25 x 366.3124 = ( 250 +31.25) ω

ω = 40.7 rad / s .

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Answer:

The answer is below

Explanation:

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h=\frac{u^2sin^2\theta}{2g}

Given that the two projectile has the same height.

For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}

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3 years ago
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Arlecino [84]
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Answer:

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PLEASEEEEEE HELP
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Explanation:

Assume east is +x.  Given:

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