The rate at which the acceleration changes.
Hope this helped.
In order to determine which ball has a higher kinetic energy, we first look at the formula of kinetic energy which is:
kinetic energy = 1/2 * mass * velocity²
The factors influencing the kinetic energy of an object are its mass and its velocity. Both of the balls have the same velocity; however, the red ball has a greater mass. Therefore,
The red ball has greater kinetic energy, because it has a greater mass.
We may also determine these quantitatively using the formula given and the kinetic energy of the yellow ball works out to be 9 J and that of the red ball is 13.5 J.
If you're referring to the different colors that usually occur at the tip of missles, rockets and some other aircraft, it either a) signifies the end of a particular plate of metal, fabricated specifically to be for the nose. Sometimes these can even be a different alloy or metal all together. or b) this shows where the curved surface begins, so in the case of damage or imperfections due to wear, they can be repaired and measured more easily. The shape of the nose is extremely important for smooth flight, and a dent or bump formed on it can make the aircraft unstable. If you can measure from where the curve starts by the difference in color, it makes repairing or re-fabricating the part much easier. Many of these curves aren't as simple as they appear.
Answer:
= Approximately 6
Explanation:
<u>What we need to know:</u>
1. Given that this is a parallel circuit, each pathway will receive the full voltage from the source. Therefore, each pathway will receive 17V.
2. The current (I) stays the same for all loads in series. Therefore, I2 and I3 are equal since they are connected in series.
3. Ohm's law states that Voltage = Current × Resistance (V=IR)(I=V/R)(R=V/I)
<u>1) Calculate the current for the path containing R2 and R3</u>
![I_2_,_3=\frac{V_2_,_3}{R_2_,_3}\\I_2_,_3=\frac{17}{5+9}\\I_2_,_3=\frac{17}{14}](https://tex.z-dn.net/?f=I_2_%2C_3%3D%5Cfrac%7BV_2_%2C_3%7D%7BR_2_%2C_3%7D%5C%5CI_2_%2C_3%3D%5Cfrac%7B17%7D%7B5%2B9%7D%5C%5CI_2_%2C_3%3D%5Cfrac%7B17%7D%7B14%7D)
Because the current running through this path is
A, then I2 is
A.
<u>2) Use Ohm's Law by plugging in I2 and R2</u>
![V_2=I_2*R_2\\V_2=\frac{17}{14} *5](https://tex.z-dn.net/?f=V_2%3DI_2%2AR_2%5C%5CV_2%3D%5Cfrac%7B17%7D%7B14%7D%20%2A5)
(approximate)
Therefore, V2 is approximately 6V.
I hope this helps!
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.