Answer:
r₂ = 0.2 m
Explanation:
given,
distance = 20 m
sound of average whisper = 30 dB
distance moved closer = ?
new frequency = 80 dB
using formula

I₀ = 10⁻¹² W/m²
now,



to hear the whisper sound = 80 dB



we know intensity of sound is inversely proportional to square of distances



r₂ = 0.2 m
From the geometry of the problem, the 20 m-long cable creates
the hypotenuse of a right triangle, with the extended of the other two sides of
size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased
by 20 m - 17.3 m = 2.7 m.
The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 *
1.6 m , or about 37044 joules.
Diagram B .... light shines through at an angle
Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x

⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.
Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2
v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared
vf^2=v0^2 + (2)(g)(yf-y0)
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m)
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2)
vf^2=108.18 m^2/s^2
vf=sqrt(108.18 m^2/s^2)
vf=10.4 m/s