When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.
The part of the light that continues on into the new medium is <em>transmitted</em>
light. Its forward progress at any point in its journey is <em>transmission</em>.
Its direction usually changes as it crosses the boundary. The bending is <em>
refraction</em>.
The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light. The process of bouncing is <em>reflection</em>.
Answer:
so rate constant is 4.00 x 10^-4 
Explanation:
Given data
first-order reactions
85% of a sample
changes to propene t = 79.0 min
to find out
rate constant
solution
we know that
first order reaction are
ln [A]/[A]0 = -kt
here [A]0 = 1 and (85%) = 0.85 has change to propene
so that [A] = 1 - 0.85 = 0.15.
that why
[A] / [A]0= 0.15 / 1
[A] / [A]0 = 0.15
here t = (79) × (60s/min) = 4740 s
so
k = - {ln[A]/[A]0} / t
k = -ln 0.15 / 4740
k = 4.00 x 10^-4
so rate constant is 4.00 x 10^-4
Answer:
6 cm long
Explanation:
F = 4110N
Vo(speed of sound) = 344m/s
Mass = 7.25g = 0.00725kg
L = 62.0cm = 0.62m
Speed of a wave in string is
V = √(F / μ)
V = speed of the wave
F = force of tension acting on the string
μ = mass per unit density
F(n) = n (v / 2L)
L = string length
μ = mass / length
μ = 0.00725 / 0.62
μ = 0.0116 ≅ 0.0117kg/m
V = √(F / μ)
V = √(4110 / 0.0117)
v = 592.69m/s
Second overtone n = 3 since it's the third harmonic
F(n) = n * (v / 2L)
F₃ = 3 * [592.69 / (2 * 0.62)
F₃ = 1778.07 / 1.24 = 1433.927Hz
The frequency for standing wave in a stopped pipe
f = n (v / 4L)
Since it's the first fundamental, n = 1
1433.93 = 344 / 4L
4L = 344 / 1433.93
4L = 0.2399
L = 0.0599
L = 0.06cm
L = 6cm
The pipe should be 6 cm long
