SOMEONE HELP ME PLEASE

Mathematics

1 answer:

Yuki888 [10]3 years ago

6 0

Answer:

1. 120 2. 45 3. 13 4. 492 5. 0.88

Step-by-step explanation:

1. set it up as a ratio 30m/3g=x/12g, cross multiply

3x=360 solve for x

x=120m

2. set it up as a ratio $40/8h=x/9h, cross multiply

8x=360 solve for x

x=$45

3. set it up as a ratio $32.5/5t=x/2t, cross multiply

5x=65 solve for x

x=$13

4. set it up as a ratio 738k/6m=x/4m, cross multiply

6x=2952 solve for x

x=492k

5. set it up as a ratio $0.32/4m=x/11m, cross multiply

4x=3.52

x=$0.88

hope this helps!

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A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 255 kg · m2 and is rotating at 9.0 rev/min about a

Degger [83]

Answer:

7 rpm = 0.73 rad/s

Step-by-step explanation:

R = Radius of merry-go-round = 1.8 m

$I_M$ = Moment of inertia of merry-go-round = 255 kg m²

$I_C$ = Moment of inertia of child

ω = 9 rev/min

m = Mass of child = 24 kg

From the conservation of angular momentum

$I\omega=I'\omega '\\\Rightarrow I\omega=(I_M+I_C)\omega'\\\Rightarrow \omega'= \frac{I\omega}{(I_M+I_C)}\\\Rightarrow \omega'=\frac{I\omega}{(I_M+mR^2)}\\\Rightarrow \omega'=\frac{255\times 9}{(255+24\times 1.8^2)}\\\Rightarrow \omega'=6.9\ rev/min$