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GaryK [48]
2 years ago
9

What does an alpha particle consist of? one proton and two neutrons two protons and two neutrons two protons and one electron on

e proton and two electrons
Physics
1 answer:
kolbaska11 [484]2 years ago
3 0

They are helium nuclei, which consist of two protons and two neutrons. The net spin on an alpha particle is zero. They result from large, perilous atoms via a process called alpha decay.

<h3>What is helium nuclei?</h3>
  • The nucleus of the helium atom also understood as the α-particle, includes two protons and two neutrons, encompassed by two electrons.
  • Alpha particles are helium nuclei with two protons and two neutrons attached. The development of their high mass and an electrical charge is their inability to infiltrate as deep as other particles such as protons and electrons.
  • Particle beams contain α (alpha)-particles, β (beta)-particles, neutron beams, etc. α-particles are helium middles consisting of two protons and two neutrons that have lived removed at high speed, while β-particles are electrons removed from a nucleus. Particle shafts also include neutron beams and proton beams.

To learn more about helium nuclei, refer to:

brainly.com/question/26226232

#SPJ4

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Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
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a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
3 years ago
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