Explanation:
The deeper the sediment layer above bedrock, the more soft soil there is for the seismic waves to travel through. Soft soil means bigger waves and stronger amplification. The earthquake damage to this building may have been influenced by the type of soil it's sitting on.
Answer:
16.53 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 18.0 m/s.
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
The maximum height reached by the ball can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 18² – (2 × 9.8 × h)
0 = 324 – 19.6h
Rearrange
19.6h = 324
Divide both side by 19.6
h = 324 / 19.6
h = 16.53 m
Therefore, the maximum height reached by the ball is 16.53 m
B: <span>It contains all of the wavelengths of the visible light spectrum.
Hope this helps mate =)</span>
Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N
Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg
Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.
For constant acceleration we have:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = -0.44 m/s^2. So we have
0 = 5.9 - 0.44t
which gives t = 5.9 / 0.44 = 13.409 s.
Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m
<span>Answer: </span>40 meters
