Answer:
Explanation:
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E = 10⁵ J
given,
Power, P = 100 TW
= 100 x 10¹² W
time, t = 1 ns
= 1 x 10⁻⁹ s
The energy of a single pulse is:-
Energy = Power x time
E = P t
E = 100 x 10¹² x 1 x 10⁻⁹
The energy contained in a single pulse is equal to 10⁵ J
Re=160ohm
Step#1
Rt=R1+R2 ( because both are in series)
Rt=(100+220 ) ohm
Rt=320 ohm
Step#2
Rt and R3 are parallel so,
Re= (Rt× R3) ÷ (Rt+R3)
Re= (320×320)÷( 320+320)
Re = 102,400÷ 640
Answer: vf = 51 m/s
d = 112 m
Explanation: Solution attached:
To find vf we use acceleration equation:
a = vf - vi / t
Derive to find vf
vf = at + vi
Substitute the values
vf = 3.5 m/s² ( 8.0 s) + 23 m/s
= 51 m/s
To solve for distance we use
d = (∆v)² / 2a
= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)
= (28 m/s)² / 7 m/s²
= 784 m/s / 7 m/s²
= 112 m
Total energy saving will be 0.8 KWH
We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW
30 bulbs are of power 60 W
So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW
Total power of 80 bulbs = 1.8+5 = 6.8 KW
Total time = 3 hour
We know that energy
Now power of each CFL bulb = 25 W
So power of 80 bulbs = 80×25 = 2000 W = 2 KW
Energy of 80 bulbs = 2×3 = 6 KWH
So total energy saving = 6.8-6 = 0.8 KWH