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2 years ago

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a. A 65-cm-diameter cyclotron uses a 500 V oscillating potential difference between the dees. What is the maximum kinetic energy

of a proton if the magnetic field strength is 0.75 T? b. How many revolutions does the proton make before leaving the cyclotron?

1 answer:

Sati [7]2 years ago

8 0

Explanation:

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A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass

Gre4nikov [31]

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

$\omega = \frac{v}{R}$

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

$KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2$

Where I is the moment of inertia previously defined.

$KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2$

In the case of the final kinetic energy, we have to,

$KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$ , then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

$1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)$

$2.2388= (\frac{7}{5}) (v)^2$

$v=1.26m/s$

7 0

3 years ago

The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from

kirill [66]

There are many ways to solve this but I prefer to use the energy method. Calculate the potential energy using the point then from Potential Energy convert to Kinetic Energy at each points.

PE = KE

From the given points (h1 = 45, h2 = 16, h<span>3 </span>= 26)

Let’s use the formula:

v2= sqrt[2*Gravity*h1] where the gravity is equal to 9.81m/s2

v3= sqrt[2*Gravity*(h1 - h3 )] where the gravity is equal to 9.81m/s2

v4= sqrt[2*Gravity*(h1 – h2)] where the gravity is equal to 9.81m/s2

Solve for v2

v2= sqrt[2*Gravity*h1]

= √2*9.81m/s2*45m

v2= 29.71m/s

v3= sqrt[2*Gravity*(h1 - h3 )

=√2*9.81m/s2*(45-26)

=√2*9.81m/s2*19

v3=19.31m/s

v4= sqrt[2*Gravity*(h1 – h2)]

=√2*9.81m/s2*(45-16)

=√2*9.81m/s2*(29)

v4=23.85m/s

7 0

2 years ago

What is an object velocity?

Blababa [14]

It is the speed with direction. (Ex. The car is going 4 mph south)

8 0

2 years ago

What happens in a change in temperature as it is heated

Andre45 [30]

It increases in temperature

5 0

2 years ago

The power is an electrical circuit is given by the equation P = i^2R, where I is the current flowing through the circuit and R i

svp [43]

Formula

Power = I^2 * R

Givens

R = 18 ohms

I = 0.1 amperes.

Solution

P = 0.1^2 * 18

P = 0.01 * 18

P = 0.18 watts.

D <<<< Answer

5 0

3 years ago

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