A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicula
1 answer:
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Explanation:
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Answer:
6005803.83105 short tons
Explanation:
The definition of density is , and the volume of a pyramid is (confusingly written on the proposal)
, so we can write:
Where s is the side of the base, being the area of that square.
We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that , and we can use this factor to convert anything written in cm to anything written in m. Example:
Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.
We can use this tool like this:
Where we have used the fact that (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.
Substituting then our values:
And now we will convert to short tons using two conversion factors at the same time:
Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
Therefore, the drill's angular displacement during that time interval is 24.17 rad.