Answer:
KE = KE (incidental) - KE of emitted photons
or KE = h * f - Wf
So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy
If you double the frequency then h * f = 6.16
KE = 6.16 - 1.2 = 4.96 eV
Answer:
Substance X has the highest temperature because its particles have more kinetic energy.
Explanation:
The temperature of a substance is directly proportional to the average kinetic energy of its particles, according to the equation:
where
is the average kinetic energy of the particles
k is the Boltzmann's constant
T is the absolute temperature of the substance
In this problem, we see that the particles in the two substances have same mass, but particles of substance X travel faster than particles of substance Y, and since the average kinetic energy depends on the square of the speed of hte particles:
we can conclude that particles in substance X have more kinetic energy than particles in substance Y, therefore substance X has higher temperature than substance Y.
The question is incomplete. The complete question is :
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .
Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:
1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
Solution :
Given :
d = 10 mm
= 0.010 m
Then, Capacitance,
Now,
And
In parallel combination,
Then energy,
b). Now the charge on the is :
Now when the capacitor gets disconnected from battery and the is slowly of the way out of the is :
Without the dielectric,
Doubling the size of a load resistor increases the load current. Increasing the load resistance, in turn the total circuit resistance is reduced. The load current would not be half as much since when you increase the size of load resistor then load current increases.
Answer:
hahha can you arrange your question correctly