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Inessa [10]
3 years ago
12

A long electric cable is suspended above the earth and carries a current of 345 A parallel to the surface of the earth. The eart

h’s magnetic field has a value of 5.6 × 10−5 T and makes an angle of 88.2 ◦ with respect to the cable. What is the magnitude of the force f exerted on a cable of length 46.9 m due to the earth’s magnetic field? Answer in units of N
Physics
1 answer:
jok3333 [9.3K]3 years ago
7 0

Answer:

0.906 N

Explanation:

Formula for magnetic force acting on current carrying cable:

F = IBLsin(\theta)

Where I = 345A is the current in the wire, B = 5.6*10^{-5} T is the magnetic magnitude generated by Earth. L = 46.9 m is the cable length. \theta = 88.2^o is the angle between vector B and cable direction.

F = 345*5.6*10^{-5}*46.9*sin(88.2^o)

F = 0.906 N

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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77
Vikki [24]
The formula for the rotational kinetic energy is

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6 0
3 years ago
provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small
gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

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T_{simple = 0.61 s

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T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / I ) ------ let this equation 2

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I = \frac{1}{3}mD²

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we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/I) OR T = 2π√(I/mgL)

so we can use I = \frac{1}{3}mD² for moment of inertia of the rod

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length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

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so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × T_{simple

we substitute in value of T_{simple

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

 

8 0
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