Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:
1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)
1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)
<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω
