Answer:
A) Emin = eV
B) Vo = (E_light - Φ) ÷ e
Explanation:
A)
Energy of electron is the product of electron charge and the applied potential difference.
The energy of an electron in this electric field with potential difference V will be eV. Since this is the least energy that the electron must reach to break out, then the minimum energy required by this electron will be;
Emin = eV
B)
The maximum stopping potential energy is eVo,
The energy of the electron due to the light is E_light.
If the minimum energy electron must posses is Φ, then the minimum energy electron must have to reach the detectors will be equal to the energy of the light minus the maximum stopping potential energy
Φ = E_light - eVo
Therefore,
eVo = E_light - Φ
Vo = (E_light - Φ) ÷ e
S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
Answer:
The cat will have <span>36J</span> of kinetic energy.
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.
From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.
Hence P(t)=a*t^2/2+10t
Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T
Hence T=225/30=7.5
It took 7.5 seconds
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