Reading a digital thermometer is more preferable because you can just easily read the measurement from a screen, a number will just show in a screen as compared to a liquid-based thermometer where you still have to read the measurement by counting the markers in the thermometer. Hope this helps! Mark brainly please!
True, scientists often talk to each other to figure out if their results were similar and what they could have done better.
Although, talking to other scientists does have risks, other scientists could copy your work and further better it.
So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol
The mass on the spring is 0.86 kg
Explanation:
The period of a mass-spring system is given by the equation

where
m is the mass
k is the spring constant
In this problem, we have:
k = 88.7 N/m is the spring constant
The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

Now we can re-arrange the first equation to solve for the mass:

Learn more about period:
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Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)
The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)
Moment of Inertia refers to:
- the quantity expressed by the body resisting angular acceleration.
- It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)
here We note that the,
In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.
The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:
I(edge) = I (center of mass) + md^2
d be the distance from an axis through the object’s center of mass to a new axis.
I2(edge) = 1/3 (m*L^2)
learn more about moment of Inertia here:
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It Increases. I just took a quiz with the same question.