Answer:
Systematic errors.
Explanation:
The density of the aluminium was calculated by a human and this is not natural but can be due to errors in the calibration of the scale for measuring the weight or taking readings from the measuring cylinder.
Random errors are natural errors. Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment. Systematic errors are due to imprecision or problems with instruments.
If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.
The <em>gaseous state</em> of matter does that. A gas expands to take the shape and volume of whatever you put it into.
Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
![\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%2Bmg%3D%5Cfrac%7Bmv%5E2%7D%7BL%7D_%7B%7D%20%5C%5C%20v%5E2%3D%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%7D%20%5Cend%7Bgathered%7D)
For the maximum velocity of the ball at the top of the vertical circular motion,
![v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T_%7B%5Cmax%20%7D%2Bmg%29%7D%7Bm%7D%7D)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%287.5_%7B%7D%2B0.31%5Ctimes9.8%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%2810.538%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B17.34%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D4.16%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
