Only within the same technology. / / /
If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.
Answer:
KE = 1/2mv^2
KE = 1/2(24)(3^2)
KE = 12(9)
KE = 108 J
Let me know if this helps!
Answer:
h f = W + KE
Input energy equals work function plus KE of emitted electron
W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19
W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev (1ev = 1.6E-19 J)
W = (16.6 - 10.1)E-19 = 6.5E-19 J
h f = 6.5E-19 J for electrons to be emitted with zero KE
f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec (threshold)
