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Murrr4er [49]
4 years ago
12

Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. A

n alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Physics
1 answer:
lorasvet [3.4K]4 years ago
3 0

Answer:

   R = r_protón / 2

Explanation:

The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement

      F = m a

Since the magnetic force is perpendicular, the acceleration is centripetal.

       a = v² / R

       

the magnetic force is

       F = q v x B = q v B sin θ

the field and the speed are perpendicular so the sin 90 = 1

we substitute

          qv B = m v² / R

          R = q v B / m v²

in the exercise they indicate

the charge  q = 2 e

the mass     m = 4 m_protón

        R = 2e v B / 4m_protón v²

we refer the result to the movement of the proton

         R = (e v B / m_proton) 1/2

the data in parentheses correspond to the radius of the proton's orbit

         R = r_protón / 2

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In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s
statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

F = ma

m = mass

a = Acceleration

By Hooke's law force is described as

F = k\Delta x

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

k\Delta x = mg

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

k (9.5-l)=0.2*g

k (9.5-l)=0.2*9.8

For state 2 we have that with 1Kg there is an elongation of 12cm

k (12-l)= 1*g

k (12-l)= 1*9.8

We have two equations with two unknowns therefore solving for both,

k = 3.136N/cm

l = 8.877cm

In this way converting the units,

k = 3.136N/cm(\frac{100cm}{1m})

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3 0
3 years ago
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6 0
4 years ago
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed
Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

\Delta x \Delta p\geq \frac{h}{4\pi}      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}

The minimum kinetic energy is calculated by using the following formula:

k=\frac{(\Delta p)^2}{2m}       (2)

m: mass of the proton = 1.67*10^{-27}kg

k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J

in eV you have:

2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV

The kinetic energy of the proton is 1.29eV

7 0
3 years ago
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