Answer:
too much exposure to the sun's rays
Answer:
Electric potential energy at the negative terminal: ![1.92\cdot 10^{-18}J](https://tex.z-dn.net/?f=1.92%5Ccdot%2010%5E%7B-18%7DJ)
Explanation:
When a particle with charge
travels across a potential difference
, then its change in electric potential energy is
![\Delta U = q \Delta V](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20q%20%5CDelta%20V)
In this problem, we know that:
The particle is an electron, so its charge is
![q=-1.60\cdot 10^{-19}C](https://tex.z-dn.net/?f=q%3D-1.60%5Ccdot%2010%5E%7B-19%7DC)
We also know that the positive terminal is at potential
![V_+=0V](https://tex.z-dn.net/?f=V_%2B%3D0V)
While the negative terminal is at potential
![V_-=-12 V](https://tex.z-dn.net/?f=V_-%3D-12%20V)
Therefore, the potential difference (final minus initial) is
![\Delta V = -12-0 = -12 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20-12-0%20%3D%20-12%20V)
So, the change in potential energy of the electron is
![\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%28-1.6%5Ccdot%2010%5E%7B-19%7D%29%28-12%29%3D1.92%5Ccdot%2010%5E%7B-18%7DJ)
This means that the electron when it is at the negative terminal has
of energy more than when it is at the positive terminal.
Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is
![1.92\cdot 10^{-18}J](https://tex.z-dn.net/?f=1.92%5Ccdot%2010%5E%7B-18%7DJ)
Answer:
a = 2.94 m/s²
Explanation:
In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:
Unbalanced Force = Frictional Force
ma = μR = μW
ma = μmg
a = μg
where,
a = maximum acceleration for the cup not to slip = ?
μ = coefficient of static friction = 0.3
g = acceleration due to gravity = 9.8 m/s²
Therefore,
a = (0.3)(9.8 m/s²)
<u>a = 2.94 m/s²</u>
The most exact answer is 78.4J also in this kind of options we can say answer "d"