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1 year ago

What relates the torque on a playground merry-go-round to the resulting angular acceleration?

1 answer:

4 0

The rotational inertia of a merry-go-round connects the torque produced by one to the angular acceleration that results.

<h3>What does rotational inertia refer to?</h3>

Any item with the ability to rotate possesses rotational inertia. It is a scalar value that indicates how challenging it is to alter the object's rotational speed around a specific rotational axis. Similar to the role that mass plays in linear mechanics, rotational inertia plays in rotational mechanics.

<h3>What is a good illustration of rotational inertia?</h3>

Rotational inertia, also referred to as the inertia moment is the resistance of an item to a change in rotation. Rotational inertia can be felt and seen in many commonplace situations, such as when using a heavy baseball bat or pushing a large group of people on the a merry-go-round.

To know more about rotational inertia visit:

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The vaule of the g constant (the acceleration of all objects due to gravity)on earth is

True [87]

The answer is: " $\frac{9.8 m}{s^2}$ " .
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6 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2

kumpel [21]

Answer:

Explanation:

120/30=6

8 0

2 years ago

Which goes into which category?

Pavel [41]

Evidence of Chemical Change:

A car rusting

Leaves changing in October

The glow of a light bulb

A white cloudy substance occurs after mixing two substances

Burning toasts

A candle being lit

Evidence of Physical Change:

Water boiling

Yellow and blue paints mixed together

Wax melting

Crushing a rock with a hammer

Hope this helps!

Have a great day!

6 0

3 years ago

What happens if you are riding your bike and hit something (like a curb) with the front wheel

evablogger [386]

You would flip forward or to the side

6 0

3 years ago