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
Answer:
The final velocity of the second car is 57 m/s south.
Explanation:
This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.
The formula to apply is :

where ;

Given in the question that;

Apply the formula as;

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}
263700+42075=87900 + 3825v₂f
305775 =87900 + 3825v₂f
305775-87900 = 3825v₂f
217875=3825v₂f
217875/3825 =v₂f
56.96 = v₂f
<u>57 m/s = v₂f { nearest whole number}</u>
Answer:
See below
Explanation:
See attached diagram
280 km east then 190 km north
Use Pythagorean theorem to find the resultant displacement
d^2 = 280^2 + 190^2
d = 338.4 km
Angle will be arctan ( 190/280) = 34.16 °