The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.
<h3>Time of travel of the P-wave</h3>
In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.
<h3>Relationship between speed and time</h3>
v ∝ 1/t
v₁t₁ = v₂t₂
t₁/t₂ = v₂/v₁
t₁/t₂ = 0.6v₁/v₁
t₁/t₂ = 0.6
t₁ = 0.6t₂
t₁ = 0.6 x 22 mins
t₁ = 13.2 mins
Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.
Learn more about P-waves here: brainly.com/question/2552909
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Draw a circuit that contains 2 batteries, three lights in parallel and a switch that controls the whole circuit.
Answer:
<h2>
128.61 Watts</h2>
Explanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque = 
= 
I is the rotational inertia = 16kgm²
To get the angular acceleration, we will use the formula;
Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts
Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
