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IRISSAK [1]
2 years ago
15

A boat with a horizontal tow rope pulls a waterskier. She skis off to the side, the rope makes an angle of 15° with the forward

direction of motion. If the tension in the rope is 180 N, how much work does the rope two on the scare during the forward displacement of 300 m?
Physics
1 answer:
dsp732 years ago
4 0

Answer:

<h3>13,976.23Joules</h3>

Explanation:

Workdone by the rope is expressed using the formula;

W = Fd sin(theta)

F is the tension in the rope = 180

d is the displacement = 300m

theta is the angle of inclination = 15°

Substitute the given parameters into the formula;

W = 180(300)sin15

W = 54000sin 15

W = 13,976.23

Hence the workdone by the rope is 13,976.23Joules

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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:

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Where;

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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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B = magnetic field

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;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

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q = charge of a proton = 1.6 x 10⁻¹⁹C

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;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

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

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

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