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Svetlanka [38]
3 years ago
8

Superconducting cables conduct current with no resistance. Consequently, it is possible to pass huge currents through the cables

, which, in turn, can produce very large forces. Two straight, parallel superconducting cables 4.5 mm apart (between centers) carry equal currents of 15000 A in opposite directions. Find the magnitude and direction of the force per unit length exerted by one conductor on the other. Should we be concerned about the mechanical strength of these wires
Physics
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

Explanation:

Force per unit length between two cables at distance d carrying current I₁ and I₂ can be given by the following expression .

F = 10⁻⁷ x 2I₁ I₂ / d

I₁ = I₂ = 15000 A

F = 10⁻⁷ x 2I₁ I₂ / d

= 10⁻⁷ x 2 x 15000² / 4.5 x 10⁻³ m

= 10000 N .

It will be repelling force ie they will repel each other because current is in opposite direction .

No , we should not be concerned about mechanical strength because force does not depend on it .

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two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma
nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force =  3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees  

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos\theta

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

 35 =√ 9x² + 25x² + 15x²    (cos 60° = 0.5)

35 = √49 x²

 x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0
3 years ago
A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5
podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity (\omega_o) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity  (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

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QveST [7]

Answer:

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