The tension on the wire is 52.02 N.
From the question, we have
Density of aluminum = 2700 kg/m3
Area,
A = πd²/4
A = π x (4.6 x 10⁻³)²/4
A = 1.66 x 10⁻⁵ m²
μ = Mass per unit length of the wire
μ = ρA
μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²
μ = 0.045 kg/m
Tension on the wire = √T/μ
34 = √T/0.045
34² = T/0.045
T = 52.02 N
The tension on the wire is 52.02 N.
Complete question:
The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.
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Answer:
Exercise 1;
The centripetal acceleration is approximately 94.52 m/s²
Explanation:
1) The given parameters are;
The diameter of the circle = 8 cm = 0.08 m
The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m
The speed of motion = 7 km/h = 1.944444 m/s
The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²
The centripetal acceleration ≈ 94.52 m/s²
The 3rd Quadrant is the lower left one on a graph,
where ' x ' and ' y ' are both negative.
Your picture is completely unrelated to this question.
solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
Answer:
C. dismiss them.
Explanation:
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