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:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s
The spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action
By the use of Ohm's law, the resistance of the bulb is 121 ohms
<h3>What is electrical power?</h3>
Let us recall that power is the rate of doing work. Now we know that the voltage of the power source is 110-V and the power is 100-W.
Since;
P= IV
I = P/V
I = 100/110
I = 0.91 A
Again;
V= IR
R = V/I
R = 110/0.91
R = 121 ohms
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