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
A supernova is a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.
Explanation:
F = 20N m= m1 a=10m/s²
m=m2 a=5m/s²
F = ma
<u>for the first one</u><u>:</u><u> </u>
f=m1 × a
20 = m1 ×10
20=10m1
m1=20/10
m1=2
<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>
f=m2×a
20=m2×5
m2= 20/5
m2= 4
since F=ma
F=(m1+m2) ×a
F =(4+2)×a
F =6×a
F=20(from the question above )
20=6×a
a=20/6
a=3.33
