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3 years ago

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What is the minimum time that a warm up should last?

1 answer:

Alchen [17]3 years ago

5 0

<h3> The minimum that a warm up should last is 5 minutes. </h3>

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Rock composed of many thin layers. This image appears in an Earth science magazine with this caption: "A non-foliated rock found

blondinia [14]

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its B

Explanation:

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3 years ago

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The wheels of a wagon can be approximated as the combination of a thin outer hoop of radius h=0.156 m and mass 4.89 kg, and two

11Alexandr11 [23.1K]

The radius of the disk of the <u>new wheels</u> should be 0.142 m. It was calculated from the moment of inertia of the <u>old wheels</u>, which is a combination of a thin outer hoop (<em>radius </em>= 0.156 m and <em>mass </em>= 4.89 kg) and two thin crossed rods (<em>mass </em>= 7.80 kg each one), and from the density (7370 kg/m³) and thickness (5.25 cm) of the <em>disk</em>.

The moment of inertia of the <u>old wheel</u> ( $I_{o}$ ) is given by the sum of the <em>moment </em>of <em>inertia </em>of the <u>outer hoop</u> and the <u>two rods</u>, as follows:

$I_{o} = I_{h} + 2*I_{r}$

$I_{o} = m_{h}r_{h}^{2} + 2*\frac{1}{12}m_{r}L_{r}^{2}$ (1)

Where:

$I_{h}$ : is the moment of inertia of the outer hoop

$I_{r}$ : is the moment of inertia of the rods

$m_{h}$ : is the mass of the hoop = 4.89 kg

$r_{h}$ : is the radius of the hoop = 0.156 m

$m_{r}$ : is the mass of each rod = 7.80 kg

$L_{r}$ : is the length of each rods = $2*r_{h}$ = 2*0.156 m = 0.312 m

Then, the <em>moment </em>of <em>inertia </em>of the <u>old wheel</u> is (eq 1):

$I_{o} = 4.89 kg*(0.156 m)^{2} + 2*\frac{1}{12}*7.80 kg*(0.312 m)^{2} = 0.246 kg*m^{2}$

The <em>moment </em>of <em>inertia </em>of the <u>new wheel</u> ( $I_{n}$ ) is the same as the <u>old wheel</u>, so:

$I_{n} = I_{o}$

$\frac{1}{2}m_{d}r_{d}^{2} = 0.246 kg*m^{2}$ (2)

The mass of the <em>disk</em>, can be calcualted from its density

$m_{d} = d*V_{d}$ (3)

The volume of the <em>disk </em>is equal to a cylinder's volume:

$V_{d} = \pi r_{d}^{2} h$ (4)

Where <em>h</em> is the thickness of the disk = 0.0525 m

By entering equations (3) and (4) into (2), we have:

$\frac{1}{2}d*V_{d}*r_{d}^{2} = 0.246 kg*m^{2}$

$\frac{1}{2}d(\pi r_{d}^{2}*h)r_{d}^{2} = 0.246 kg*m^{2}$

$\frac{1}{2}d*\pi*h*r_{d}^{4} = 0.246 kg*m^{2}$

$r_{d} = (\frac{2*0.246 kg*m^{2}}{\pi*d*h})^{1/4} = (\frac{2*0.246 kg*m^{2}}{\pi*7370 kg/m^{3}*0.0525 m})^{1/4} = 0.142 m$

Therefore, the radius of the disk should be 0.142 m.

Find more here:

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I hope it helps you!

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3 years ago

Which of the following is a FALSE statement? A. Heat moves through solids by conduction. B. Molecules move faster in warmer subs

matrenka [14]

C is the FALSE statement.

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1. What is the gravitational force between two 4 kilogram masses separated by a distance of 5

Anon25 [30]

Yo, I assumed that this situation occurs on Earth otherwise the answer would be different.

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Uranus has more than 14 times as much mass as earth, yet the gravitational force is less. how can that be?

Anarel [89]

It has more mass, yes. But it has less of a gravitational pull because it is farther away from the sun than the Earth is

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4 years ago

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