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Nesterboy [21]
3 years ago
8

In the amusement park ride Mr. Freeze, riders are uniformly accelerated from rest by magnetic induction motors along a 70 meter

horizontal track in just 5 seconds. While accelerating, friction exerts 500N of force on the train. Then the train coasts through the loops and turns of the remainder of the ride. A train is loaded with passengers so that it has a total mass of 2500 kg. Determine the acceleration. Net Force, and all of the Forces acting upon the train during the ride knowing that forces are vectors.
Physics
1 answer:
pishuonlain [190]3 years ago
7 0

Answer:

Acceleration=5.6 m/s

Explanation:

Dunno the rest

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
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Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

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We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

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θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

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v = \frac{qBr}{m}

From the question;

B = 1.25T

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r = 0.40m

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Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

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K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

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