Answer:
Explanation:
In the SQL database manipulation coding language, there are a variety of different available data types to use. These data types are the following.
- Numeric
- Date/Time
- Character/String
- Unicode Character/String
- Binary
- 3rd Party / Miscellaneous
Each of these data types has a wide range of subcategories (data types as well) that each handle different variables in a coding sequence. These subcategories can be all seen in the attached picture below.
Answer:
A. 0.0450
B. 4
C. 0.25
D. 37.68
E. 6Hz
F. -0.523
G. 1.5m/s
H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)
I. -1.67m/s.
Explanation:
Given the equation:
y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)
Standard wave equation:
y(x, t)=Asin(kx−ωt+ϕ)
a.) Amplitude = 0.0450
b.) Wave number = 1/ λ
λ=2π/k
From the equation k = 25.12
Wavelength(λ ) = 2π/25.12 = 0.25
Wave number (1/0.25) = 4
c.) Wavelength(λ ) = 2π/25.12 = 0.25
d.) Angular frequency(ω)
ωt = 37.68t
ω = 37.68
E.) Frequency (f)
ω = 2πf
f = ω/2π
f = 37.68/6.28
f = 6Hz
f.) Phase angle(ϕ) = -0.523
g.) Wave propagation speed :
ω/k=37.68/25.12=1.5m/s
h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)
(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.
Doesn't look like there are any..
Answer:
what do you need help with
