All categories

2 years ago

What is the relationship between mechanical stress inside of a cylindrical part with a given force applied to it

1 answer:

7 0

The relationship between mechanical stress inside of a cylindrical part with a given force applied to it is stress = force/area.

<h3>How is pressure associated with strain?</h3>

The utility of pressure offers upward thrust to strain, however now no longer the opposite manner around. A pressure is a push or pull upon an item attributable to the item's interplay with some other item while Stress is an inner resistance to supplied through the frame itself each time it's far beneath a few type of deformation.

Stress is denoted through σ. It is represented as N/m2. Stress formulation is made use of to locate carried out on any given frame if pressure and region on which pressure is exerted is given withinside the problem.

You might be interested in

Why must air tanks be drained

Jobisdone [24]

Water can freeze in cold weather and cause brake failure.

7 0

3 years ago

About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa

Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

$K.E = \frac{2KZe^2}{r}$

where;

Z is the atomic number of aluminium = 13
e is charge
r is distance of closest approach = thickness of aluminium
k is Coulomb's constant = 9 x 10⁹ Nm²/C²

<h3>For 2.5 MeV electrons</h3>

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

7 0

2 years ago

5. Which of the following is false about onStep?

katovenus [111]

The false statement about onStep is: B. The default number of steps per second is 30.

<h3>What is an onStep?</h3>

An onStep can be defined as a computerized telescope goto controller that is designed and developed to <u>animate shapes</u> while using it on a variety of mounting systems such as forks.

<h3>The characteristics of an onStep.</h3>

In Engineering, some of the characteristics that are associated with an onStep include the following:

The onStep function can be called without user input.

It can be used to animate shapes without user input.

It only runs a certain number of times.

In conclusion, the default number of steps per second for onStep isn't 30.

Read more on onStep here: brainly.com/question/25619349

7 0

2 years ago

PLEASE HELP QUICK!!

ivolga24 [154]

R01= 14.1 Ω

R02= 0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω