For resistance we have R=ρ l/a
thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ
σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
=0.26 Ω-1
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
Time taken to reach water :
Now, initial vertical speed , u = 0 m/s.
By equation of motion :
Here, a = g = acceleration due to gravity = 9.8 m/s².
So,
Therefore, the height of the bridge is 3.46 m.
Hence, this is the required solution.
