Answer:
S = V0 t + 1/2 a t^2
S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)
S = 1500 m + .6 * 90000 m = 55,500 m
Check: V0 = 5 m/s
V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s
Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)
S = 185 * 300 = 55,500 m
We calculated V2 above at 365 m/s the speed after 300 sec
Cause he left out the noble gases out of the periodic table for one good reason, 1: He did not know them
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
3 times louder
Explanation:
The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².
Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹
and I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₁ = 10㏒(I₁/I₀)
L₁ = 10㏒(10⁻⁹/10⁻¹²)
L₁ = 10㏒(10³)
L₁ = 3 × 10㏒10
L₁ = 30㏒10
L₁ = 30 dB
Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₂ = 10㏒(I₁/I₀)
L₂ = 10㏒(10⁻³/10⁻¹²)
L₂ = 10㏒(10⁹)
L₂ = 9 × 10㏒10
L₂ =90㏒10
L₂ = 90 dB
So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3
So, Braylee's music is 3 times louder than Jessica's music
