4 Ω Ri
10 Ω
R4
ΑΛΛ-
5V
+
R6
R2
4 Ω
2 Ω
R5
4 Ω
R7
R3
8 Ω
2 Ω
Your answer

Ibelieve that the correct answer is: c. along plate boundaries

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Answer(b) both involve the use of levers;

Amicroscope, because these cells are so small that you would not be able to see them, or what they attack without it.