Answer:
Phylum Annelida commonly referred as segmented worms possess long , cylindrical and segmented body .
Phylum Aschelminthes commonly referred as round worms possess long , cylindrical , unsegmented body and show sexual dimorphism .
Phylum Echinodermata which includes star fish have tube feet as locomotory organ .
Phylum Porifera commonly referred as pore bearing animals and are diploplastic which includes euspongia etc.
Answer
Given,
y(x, t) = (3.5 cm) cos(2.7 x − 92 t)
comparing the given equation with general equation
y(x,t) = A cos(k x - ω t)
A = 3.5 cm , k = 2.7 rad/m , ω = 92 rad/s
we know,
a) ω =2πf
f = 92/ 2π
f = 14.64 Hz
b) Wavelength of the wave
we now, k = 2π/λ
2π/λ = 2.7
λ = 2 π/2.7
λ = 2.33 m
c) Speed of wave
v = ν λ
v = 14.64 x 2.33
v = 34.11 m/s
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
radio waves bc they have the longest wave lenthgs in a magnetic spectrum
