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MrRissso [65]
3 years ago
7

what is the resistance of a 12m long wire having radius( 2*10 to the power - 4) and resistivity (3.14*10 to the power - 8)ohmM

Physics
1 answer:
Arisa [49]3 years ago
3 0
R = ρl/A

Where R = Resistance in Ohms, Ω, ρ = Resistivity in Ωm, l = Length in m.

Area in m²

ρ = Resistivity = 3.14 * 10⁻⁸ Ωm,  Length l = 12m, 

Area = πr² = π* (2*10⁻⁴)² m² ≈ 3.14 * (2*10⁻⁴)² m²


R = ρl/A

    ≈ 3.14 * 10⁻⁸ * 12 / (3.14 * (2*10⁻⁴)²)

     ≈ 3

Resistance, R ≈ 3 Ω
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Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

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2.40 x 10⁻¹³ C

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n_{e} = number of electrons = 6.25 x 10⁶

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6 0
3 years ago
A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
3 years ago
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