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3 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzz help 20 points

Physics

2 answers:

iris [78.8K]3 years ago

7 0

${\underline{\pink{\textsf{\textbf{ Answer : }}}}}$

➩ 1.23 feet

${\underline{\purple{\textsf{\textbf{Explanation : }}}}}$

Given :

Simon cuts a pipe that was 4.92 feet long
Then he cuts it into four equal pieces.

To find :

What is the length of the each piece.

Solution :

As it is told that it's divided into four equal pieces

Therefore,

We must divide it by 4 to get the length of each piece.

So,

$\sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.$

kakasveta [241]3 years ago

5 0

Answer:

1.23

Explanation:

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3 years ago

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The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

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${}$ ⇩

[m = 20 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R]

${}$ ⇩

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${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

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Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²· $\omega _i$ = (0.5·M·R² + m·r²)· $\omega _f$

Given that 0.5·M·R²· $\omega _i$ is constant, as the value of m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

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Answer:

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4 0

3 years ago