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Answer:
The ball will drop 0.881 m by the time it reaches the catcher.
Explanation:
The position of the ball at time "t" is described by the position vector "r":
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
When the ball reaches the catcher, the position vector will be "r final" (see attached figure).
The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.
Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):
x = x0 + v0x · t
17.0 m = 0 m + 40.1 m/s · t
t = 17.0 m/ 40. 1 m/s = 0.424 s
With this time, we can calculate the y-component of the vector "r final", the drop of the ball:
y = y0 + v0y · t + 1/2 · g · t²
Initially, there is no vertical velocity, then, v0y = 0.
y = 1/2 · g · t²
y = -1/2 · 9.8 m/s² · (0.424 s)²
y = -0.881 m
The ball will drop 0.881 m by the time it reaches the catcher.
Answer:
Answer:
β = 114 db
Explanation:
The intensity of sound in decibles is
β = 10 log
in most cases Io is the hearing threshold 1 10-12 W / cm²
let's calculate the intensity of each instrument
I / I₀ = 10 (β / 10)
I = I₀ 10 (β / 10)
trumpet
I1 = 1 10⁻¹² 10 (94/10)
I1 = 2.51 10⁻³ / cm²
Thrombus
I2 = 1 10⁻¹² 10 (107/10)
I2 = 5.01 10-2 W / cm²
low
I3 =1 1-12 (113/10) W/cm²
I3 = 1,995 10-1 W / cm²
when we place the three instruments together their sounds reinforce
I_total = I₁ + I₂ + I₃
I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1
I_total = 0.00251 + 0.0501 + 0.1995
I_total = 0.25211 W / cm²
let's bring this amount to the SI system
β = 10 log (0.25211 / 1 10⁻¹²)
β = 114 db