Answer:
Initial Velocity is 4 m/s
Explanation:
What is acceleration?
It is the change in velocity with respect to time, or the rate of change of velocity.
We can write this as:
![a=\frac{\Delta v}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D)
Where
a is the acceleration
v is velocity
t is time
is "change in"
For this problem , we are given
a = 1.2
t = 10
Putting into formula, we get:
![a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D%5C%5C1.2%3D%5Cfrac%7B%5CDelta%20v%7D%7B10%7D%5C%5C%5CDelta%20v%20%3D%201.2%2A10%5C%5C%5CDelta%20v%20%3D%2012)
So, the change in velocity is 12 m/s
The change in velocity can also be written as:
![\Delta v = Final \ Velocity - Initial \ Velocity](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20Final%20%20%5C%20Velocity%20-%20Initial%20%5C%20Velocity)
It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:
![\Delta v = Final \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4](https://tex.z-dn.net/?f=%5CDelta%20v%20%3D%20Final%20%20%5C%20Velocity%20-%20Initial%20%5C%20Velocity%5C%5C12%3D16-Initial%20%5C%20Velocity%5C%5CInitial%20%5C%20Velocity%20%3D%2016%20-%2012%20%3D%204)
hence,
Initial Velocity is 4 m/s
Answer:
Explanation:
The concept of a new strong nuclear force was introduced. In 1935, the first theory for this new force was developed by the Japanese physicist Hideki Yukawa, who suggested that the nucleons would exchange particles between each other and this mechanism would create the force.
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As
![V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC](https://tex.z-dn.net/?f=V_%7Bpotential%7D%3Dkq%2Fr%5C%5C%20q%3DrV_%7Bpotential%7D%2Fk%5C%5Cq%3D%5Cfrac%7B%280.20%29%28240%29%7D%7B9%2A10%5E%7B9%7D%20%7D%5C%5C%20q%3D5.333%2A10%5E%7B-9%7DC%5C%5Cor%5C%5C%20q%3D5.33nC)
For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Iron III ion is an iron ion with a +3 charge. If iron bonded with oxygen, it would form Fe2O3 which is rust. In this case, the oxidation number on Fe is +3 and it is -2 on oxygen. This would be called iron (III) oxide.
<span>The oxidation number of an ion times the number of the ions must equal zero when added together in a molecule with no charge. In this case, there are 2 iron molecules and each has a positive 3 charge. 2x3=6. There are three oxygen molecules each with a negative 2 charge. 3x-2=-6. 6+-6=0</span>