If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.
If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.
I think it’s b... not sure tho sorry
Answer:
A) 199.78 J
B) 9.292x10^14 J
C) 4.2x10^7 m/s
D) 0.65 m
E) 1.13x10^-8 sec
D) 2.94x10^-9 sec
Explanation:
mass of ball = 0.0580 kg
A)
If smashed at v = 83.0 m/s, KE is
KE = 0.5mv^2
= 0.5 x 0.0580 x 83.0^2
= 199.78 J
B) if returned at v = 1.79×10^8 m/s, KE will be
KE = 0.5mv^2
= 0.5 x 0.0580 x (1.79×10^8)^2
= 9.292x10^14 J
C) during Einstein's return, velocity of rabbit relative to players is
Vr = 2.21×108 m/s
Rabbit's velocity relative to ball = 2.21×10^8 - 1.79×10^8
= 4.2x10^7 m/s
D) the rabbit's speed approaches the speed of light so we consider relativistic effect. The rabbit's measured distance is
l = l°( 1 - v^2/c^2)
= 2.5(1 - 2.21/3)
= 2.5 x 0.26
= 0.65 m
E) according to the players, the time taken by the rabbit is
t = d/v = 2.5/ 2.21×10^8
= 1.13x10^-8 sec
F) the time for rabbit as measured by rabbit is relativistic
t = t°( 1 - v^2/c^2)
= 1.13x10^-8 (1 - 2.21/3)
= 1.13x10^-8 x 0.26
= 2.94x10^-9 sec
