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2 years ago

A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the

Mathematics

1 answer:

ira [324]2 years ago

8 0

Answer:

7 1/17

Step-by-step explanation:

A figure can be helpful.

The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.

The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...

BC² = BD² +CD²

BC² = 8² +15² = 289

BC = √289 = 17

The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...

(long side)/(hypotenuse) = DE/BD = CD/BC

DE = BD·CD/BC = 8·15/17

DE = 7 1/17 ≈ 7.059

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3 years ago

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstre

Vlad1618 [11]

Answer:r'=0.327 m

Step-by-step explanation:

Given

$N=2.85 rev/s$

angular velocity $\omega =2\pi N=17.90 rad/s$

mass of objects $m=1.5 kg$

distance of objects from stool $r_1=0.789 m$

Combined moment of inertia of stool and student $=5.53 kg.m^2$

Now student pull off his hands so as to increase its speed to 3.60 rev/s

$\omega _2=2\pi N_2$

$\omega _2=2\pi 3.6=22.62 rad/s$

Initial moment of inertia of two masses $I_0=2mr_^2$

$I_0=2\times 1.5\times (0.789)^2=1.867$

After Pulling off hands so that r' is the distance of masses from stool

$I_0'=2\times 1.5\times (r')^2$

Conserving angular momentum

$I_1\omega =I_2\omega _2$

$(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62$

$I_0'=1.397\times 0.791$

$I_0'=5.851$

$5.53+2\times 1.5\times (r')^2=5.851$

$2\times 1.5\times (r')^2=0.321$

$r'^2=0.107009$

$r'=0.327 m$

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3 years ago