It’s C
solar
correct me if i’m wrong though
As per FBD while its accelerating upwards
we can say that

here normal force is given as


now mass is given as


now we will have


Now while accelerating downwards we can say by FBD

again plug in all values


Answer:
Change in temperature ∆(tita) is 266.097°C
Explanation:
Ok kinectic energy = 1/2MV²
5.4 grams =( 5.4/1000) kilogram
Kinectic energy =( 1/2 )*(5.4/1000)*261²
Kinectic energy = 183.9267 joules
If kinetic energy = thermal energy
183.9267 joules = mc∆(tita)
Where ∆(tita) = change in temperature
And c = 128 J/kg
∆(tita) = 183.9267/((5.4/1000)*128)
∆(tita) = 266.097
∆(tita) = 266.097°C
X=.5(a)t^2 can be used: 2.5m=.5(g)(1), g=5m/s^2.
Hf = Ф + Kmax
Where,
h = 4.14*10^-15 eV.s
f = 3.0*10^15 Hz
Kmax = 7.27 eV
Ф = ?
Therefore,
Ф= hf-Kmax = 4.14*10^-5*3.0*10^15 - 7.27 = 5.15 eV