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n200080 [17]
3 years ago
11

Tyler is in downtown San Antonio entertaining the tourists with amazing feats of physics. He carries a nickel

Physics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

The velocity of the nickel just before it hits the ground is approximately 49.3 m/s

Explanation:

In order to calculate the velocity of the nickel just before it hits the ground, we recall the kinematic equation of motion, v² = u² + 2·g·h, where the variables of the equation are defined as follows;

v = The velocity of the nickel just before it hits the ground after it is pushed off the observation deck

u = The initial velocity of the nickel just before it is pushed off the observation deck = 0 m/s

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which he nickel is pushed off the edge = 124 m

Substituting the given values and the constant, "g", we have;

v² = 0² + 2 × 9.8 × 124 = 2,430.4

v = √2,430.4

Using a graphing calculator, we have;

v = √2,430.4 = 14·√(62/5) ≈ 49.3

The velocity of the nickel just before it hits the ground = v ≈ 49.3 m/s.

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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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Explanation:

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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;

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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°

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Divide both side by v;

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

Make v subject of the formula

v = \frac{qBr}{m}

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B = 1.25T

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r = 0.40m

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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