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
Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol
<h3>
<em>Al(OH)3 = 78.01 g/mol</em></h3>
Answer:
t=0.42s
Explanation:
Here you have an inelastic collision. By the conservation of the momentum you have:

m1: mass of the bullet
m2: wooden block mass
v1: velocity of the bullet
v2: velocity of the wooden block
v: velocity of bullet and wooden block after the collision.
By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

hence, the time is t=0.42 s