From everyday questions to specialized queries, IDNLearn.com has the answers. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
To solve this problem, we need to find the electric potential [tex]\( V \)[/tex] given the charge [tex]\( Q \)[/tex] and the electric potential energy [tex]\( U \)[/tex]. The relationship between these quantities is given by the formula:
[tex]\[ V = \frac{U}{Q} \][/tex]
where:
- [tex]\( U \)[/tex] is the electric potential energy,
- [tex]\( Q \)[/tex] is the charge,
- [tex]\( V \)[/tex] is the electric potential.
The given values are:
- [tex]\( Q = 7.2 \times 10^{-5} \)[/tex] coulombs,
- [tex]\( U = 1.08 \times 10^{-2} \)[/tex] joules.
Now, we substitute these values into the formula:
[tex]\[ V = \frac{1.08 \times 10^{-2}}{7.2 \times 10^{-5}} \][/tex]
Dividing these values gives:
[tex]\[ V = 150.0 \, \text{volts} \][/tex]
Rounding the electric potential to the nearest whole number:
[tex]\[ V \approx 150 \, \text{volts} \][/tex]
Thus, the electric potential, to the nearest whole number, is [tex]\( 150 \)[/tex] V.
[tex]\[ V = \frac{U}{Q} \][/tex]
where:
- [tex]\( U \)[/tex] is the electric potential energy,
- [tex]\( Q \)[/tex] is the charge,
- [tex]\( V \)[/tex] is the electric potential.
The given values are:
- [tex]\( Q = 7.2 \times 10^{-5} \)[/tex] coulombs,
- [tex]\( U = 1.08 \times 10^{-2} \)[/tex] joules.
Now, we substitute these values into the formula:
[tex]\[ V = \frac{1.08 \times 10^{-2}}{7.2 \times 10^{-5}} \][/tex]
Dividing these values gives:
[tex]\[ V = 150.0 \, \text{volts} \][/tex]
Rounding the electric potential to the nearest whole number:
[tex]\[ V \approx 150 \, \text{volts} \][/tex]
Thus, the electric potential, to the nearest whole number, is [tex]\( 150 \)[/tex] V.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions are important to us at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.