Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;
where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length
Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s
To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency
Speed of sound in air
The definition of wavelength is,
Here,
v = Velocity
f = Frequency
Replacing,
Therefore the wavelength of that tone in air at standard conditions is 0.589m