Answer:
A mid shot or medium shot
Explanation:
Ummm what is the question
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.
Answer: b) R1 will connect to a physical Ethernet link, with the other end of the cable connected to a device at the WAN service provider point of presence.
.d) R1 will forward data link frames to R2 using an Ethernet header/trailer.
Explanation:
Here we have to understand what is MLPS. MLPS is a protocol which identifies the shortest route for the transfer of messages between routers instead of the longest route.
Here we are given a WAN which is a wide area network. Using layer 2 Ethernet service the frames are transmitted across the routers within the WAN.
Option A is incorrect as as connecting the other end of the cable to the R2 would reduce the functionality of the WAN. Option C is incorrect as HDLC header trailer has no effect.
Option B and D are correct as Ethernet header/ trailer has the same size every frame and to maintain the functionality of the WAN one end must be connected to point of presence(PoP).
Answer:
b, with grace and maturity
