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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Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
Answer:
<u><em>A. wavelength</em></u>
Explanation:
The others are about sound and how high it is. That has nothing to do with time.