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seropon [69]
3 years ago
15

A car of mass 487 kg travels around a flat, circular race track of radius 53.3 m. The coefficient of static friction between the

wheels and the track is 0.19. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track? Answer in units of m/s.
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

9.96 m/s

Explanation:

mass of car, m = 487 kg

radius of track, R = 53.3 m

coefficient of static friction, μ = 0.19

acceleration due to gravity, g = 9.8 m/s^2

let v be the maximum speed so that the car can go without flying off the track.

The formula for the maximum speed is given by

v_{max}=\sqrt{\mu Rg}

v_{max}=\sqrt{0.19\times53.3\times9.8

vmax = 9.96 m/s

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A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
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Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

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current carried by the wire, I = 5.6 A

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F = BIl \ sin(\theta)

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F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

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F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

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