Answer:
Explanation:
Not really sure what you're trying to do. You propagate uncertainties for indirect measurements, as in when you calculate a value from other values.
What you have here is a series of values of direct measurements it seems.
Anyway, for error bars will have a width of 2 times the uncertainty reported.
For example on the first one
373.67 +/- 15.444
You would have an error bar with a width of 2 * 15.444 = 30.888. This bar would be centered at 373.677. The lowest point of the error bar would be at 358.233 and the highest point at 389.121.
You also mentioned a scatter plot, but scatter plots are 2D at least. Are these measurements associated to something else like time? You need 2 coordinates for each point in a scatter plot.
The name is fluorine-19.
Usually, in the names of isotopes, we say the name of the element followed the mass number (eg. Carbon-12). The mass number is the sum of the number of protons and neutrons of that atom. Electrons are not included since they're too light compared to protons and neutrons.
Therefore, to find out the name of the isotope, we have to find the mass number. Add up 9 and 10, which makes 19, so, the answer is fluorine-19.
Answer:
A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.
(Hope this helps! Btw, I am the first to answer.)
Explanation:
gas vibrate and moves freely at high speeds, liquid vibrate, move about, and slide past each other. solid vibrate (jiggle) but generally do not move from place to place
hope it helps you
Answer:
2m₁m₃g / (m₁ + m₂ + m₃)
Explanation:
I assume the figure is the one included in my answer.
Draw a free body diagram for each mass.
m₁ has a force T₁ up and m₁g down.
m₂ has a force T₁ up, T₂ down, and m₂g down.
m₃ has a force T₂ up and m₃g down.
Assume that m₁ accelerates up and m₂ and m₃ accelerate down.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₁ − T₂ − m₂g = m₂(-a)
T₁ − T₂ − m₂g = -m₂a
(m₁g + m₁a) − T₂ − m₂g = -m₂a
m₁g + m₁a + m₂a − m₂g = T₂
(m₁ − m₂)g + (m₁ + m₂)a = T₂
Sum of the forces on m₃:
∑F = ma
T₂ − m₃g = m₃(-a)
T₂ − m₃g = -m₃a
a = g − (T₂ / m₃)
Substitute:
(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂
(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂
(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂
m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂
2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂
T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)
