me ajudem por favor pra agora de noite
Part a)
per day electricity power consumed when 100 W bulb is used for 8 hours

for one year consumption

now the cost will be given

now when other energy efficient light is used

for one year consumption

now the cost will be given

Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:

where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,

<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:

(c)
Now we will consider the downward motion and use the third equation of motion:

where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,

<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:

(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
Is this like a zoom or something like that