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Alika [10]
3 years ago
15

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

length of each rod is 0.85 m, and the mass of each is 0.073 kg. One rod is held in place above the ground, while the other floats beneath it at a distance of 8.2 × 10−3 m. Determine the current in the rods.
Physics
1 answer:
bazaltina [42]3 years ago
5 0

Answer:

The current in the rods is 171.26 A.

Explanation:

Given that,

Length of rod = 0.85 m

Mass of rod = 0.073 kg

Distance d = 8.2\times10^{-3}\ m

The rods carry the same current in the same direction.

We need to calculate the current

I is the current  through each of the wires then the force per unit length on each of them is

Using formula of force

\dfrac{F}{L}=\dfrac{\mu_{0}I^2}{2\pi d}

\dfrac{mg}{L}=\dfrac{\mu_{0}I^2}{2\pi d}

Where, m = mass of rod

l = length of rod

Put the value into the formula

I^2=\dfrac{mgd}{\mu L}

I^2=\dfrac{0.073\times9.8\times8.2\times10^{-3}}{2\times10^{-7}}

I=\sqrt{29331.4}

I=171.26\ A

Hence, The current in the rods is 171.26 A.

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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
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Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

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4 years ago
Which parts of The Action Potential Are Represented On The ECG?
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<h3>Which parts of The Action Potential Are Represented On The ECG?</h3>
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When a certain air-filled parallel-plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 160 µC.
ozzi

Answer:

k = 2.45

Explanation:

It is given that,

Charge acquired on each plate of the capacitor due to a battery, q=160\ \mu C

If V is the potential difference across the plates, capacitance in air is given by :

C_a=\dfrac{q}{V}=\dfrac{160}{V}.............(1)

Let k is the dielectric constant of the dielectric slab. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 220 µC on each plate.

New charge becomes, Q=220\ \mu C+220\ \mu C=440\ \mu C

C_d=\dfrac{Q}{V}=\dfrac{440}{V}.............(2)

The dielectric constant of the dielectric slab is given by :

k=\dfrac{C_d}{C_a}

k=\dfrac{440}{160}

k = 2.45

So, the dielectric constant of the dielectric slab is 2.45. Hence, this is the required solution.

7 0
3 years ago
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