Answer:
constant velocity unless acted on my an opposite force
Acceleration is how much the velocity changes within a period of time so,
Acceleration= is the change in velocity divided by change in time
your units will be m/s squared
"Ionization energy" is the one among the following choices given in the question that <span>decreases with increasing atomic number in Group 2A. The correct option among all the options that are given in the question is the third option or option "C". I hope that the answer has helped you.</span>
Answer:
The difference is 7.6 grams.
Explanation:
In mathematics the difference of two numbers is express as the subtraction between them:

So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.
Therefore, it is get:

<u>Hence, the difference is 7.6 grams. </u>
The result of 7.628 will be expressed as 7.6 to have the correct number of significant figures.
Notice how that can be express in units of kilograms too since there is 1000 gram in 1 kilogram:
⇒ 
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ
= m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s