Answer:
v = 0.42m/s
Explanation:
In order o calculate the linear speed of the point at the border of the wheel, you first take into account that the total acceleration of such a point is given by:
(1)
atotal: total acceleration = 1.2m/s^2
ar: radial acceleration of the wheel
at: tangential acceleration
The tangential acceleration is also given by:
(2)
r: radius of the wheel = (40cm/2 )= 20cm = 0.2m
α: angular acceleration = 4.0rad/s^2
You replace the expression (2) into the expression (1) and solve for the radial acceleration:
Next, you use the following formula for the radial acceleration and solve for the linear speed:
The linear speed of the point at the border of the wheel is 0.42m/s
<span>The scientists use data
and measurement to obtain empirical evidence. Data can be collected through
direct observation or else experimentation. Empirical measurements and data can
be gathered by using qualitative and quantitative methods. Empirical evidence
contains the recording and analyzing the data which is a central part of scientific
method.</span>
Explanation:
Vi = 12 m/s
a = 3 m/s^2
t = 2 s
Vf = Vi + a × t = 12 + 3 ×2 = 18 m/s
Density of powder 1 = 0.5 g / 45 cm^3 = 1/90 g/cm^3
Density of powder 2 = 1.3 g / 65 cm^3 = 1/50 g/cm^3
Therefore the densities of the two powders are different, hence chemical reaction has occurred.
(note: none of the other choices make sense. In fact, a different density does not necessarily indicate a chemical change, see paragraph below).
Density of powders are not definitive unless they are each of the same size and texture. For example, granular sugar, rock sugar, and icing sugar all have different densities. I would conclude that this experiment does not lend to a reliable answer.
