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use polar coordinates to find the volume of the given solid. under the paraboloid z = x2 y2 and above the disk x2 y2 ≤ 25

User Zeugor
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Final answer:

The radius of curvature of a 25.0-MeV proton moving perpendicular to a 1.20-T magnetic field of a cyclotron is calculated by first converting kinetic energy to speed, then applying the formula r = mv/qB.

Step-by-step explanation:

The student is asking for the radius of curvature of the path of a 25.0-MeV proton moving perpendicularly to a 1.20-T magnetic field in a cyclotron. To calculate the radius of curvature (r), we can use the formula derived from equating the magnetic force to the centripetal force that causes the proton to move in a circular path:

r = mv/qB

where:

  • m is the mass of the proton,
  • v is the speed of the proton,
  • q is the charge of the proton, and
  • B is the magnetic field strength.

First, convert the proton's kinetic energy (KE) from MeV to joules (J), then use it to calculate the speed (v) using the formula KE = 1/2 mv^2. Finally, insert the values of m, v, q, and B into the radius of curvature formula to find the answer.

User Andrey Kryukov
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4 votes

Final answer:

The radius of curvature for a 25.0-MeV proton moving perpendicular to a 1.20-T magnetic field is found using the formula r = mv/(qB), where the mass (m) and charge (q) of the proton are constants and the velocity (v) is derived from the proton's kinetic energy.

Step-by-step explanation:

To find the radius of curvature of the path of a 25.0-MeV proton moving perpendicular to a 1.20-T magnetic field of a cyclotron, we apply the principles of circular motion and magnetic force.

The formula that relates the magnetic force exerted on a moving charge to the radius of curvature (r) of its circular path is given by:
F = qvB = mv²/r
where:
q is the charge of the proton,
v is the velocity of the proton,
B is the magnetic field strength,
m is the mass of the proton,
r is the radius of curvature.

To solve for r, rearrange the formula to get:
r = mv/(qB).

The energy (E) of the proton is given as 25.0 MeV, which can be converted to joules (J). Using the relationship E = (1/2)mv², we can solve for the velocity (v) of the proton:

v = √(2E/m)

The mass (m) and charge (q) of a proton are known constants (m = 1.6726219 × 10⁻²⁷ kg, q = 1.602176634 × 10⁻¹⁹ C).

After calculating velocity (v), substitute the values of v, m, q, and B into the radius formula to determine the radius of curvature (r) for the proton's path in the cyclotron.

User ChrisArmstrong
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