The first PCS technology used a form of time division multiplexing called ____. a. Time Division Multiple Access (TDMA) b. Gl
1 answer:
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Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.