<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer:
It would be A, because it is has more height in which the potential energy would be greater.
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
Weight = Mass * gravity
= 1470* 9.8 = 14406 N ≈ 14,400 N