Volumes of liquids such as water can be readily measured in a graduated cylinder.
Explanation:
1. Mass of an object
2. Distance between the objects
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,
![R_{eq}= \frac{V}{I}](https://tex.z-dn.net/?f=R_%7Beq%7D%3D%20%5Cfrac%7BV%7D%7BI%7D)
Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,
![P=VI \rightarrow I = \frac{P}{V}](https://tex.z-dn.net/?f=P%3DVI%20%5Crightarrow%20I%20%3D%20%5Cfrac%7BP%7D%7BV%7D)
![I =0.608 A](https://tex.z-dn.net/?f=I%20%3D0.608%20A)
Applying Ohm's law
![R_{eq} = \frac{120V}{0.608A}](https://tex.z-dn.net/?f=R_%7Beq%7D%20%3D%20%5Cfrac%7B120V%7D%7B0.608A%7D)
![R_{eq} = 197.4\Omega](https://tex.z-dn.net/?f=R_%7Beq%7D%20%3D%20197.4%5COmega)
Therefore the equivalent resistance of the light string is ![197.4\Omega](https://tex.z-dn.net/?f=197.4%5COmega)
Answer: The speed necessary for the electron to have this energy is 466462 m/s
Explanation:
Kinetic energy is the energy posessed by an object by virtue of its motion.
![K.E=\frac{1mv^2}{2}](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1mv%5E2%7D%7B2%7D)
K.E= kinetic energy = ![0.991\times 10^{-19}J](https://tex.z-dn.net/?f=0.991%5Ctimes%2010%5E%7B-19%7DJ)
m= mass of an electron = ![9.109\times 10^{-31}kg](https://tex.z-dn.net/?f=9.109%5Ctimes%2010%5E%7B-31%7Dkg)
v= velocity of object = ?
Putting in the values in the equation:
![0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}](https://tex.z-dn.net/?f=0.991%5Ctimes%2010%5E%7B-19%7DJ%3D%5Cfrac%7B1%5Ctimes%209.109%5Ctimes%2010%5E%7B-31%7Dkg%5Ctimes%20v%5E2%7D%7B2%7D)
![v=466462m/s](https://tex.z-dn.net/?f=v%3D466462m%2Fs)
The speed necessary for the electron to have this energy is 466462 m/s