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BartSMP [9]
3 years ago
14

A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable su

rface crack length if the surface energy of polystyrene is 0.50 J/m2 (2.86 103 in.-lbf/in.2 ). Assume a modulus of elasticity of 3.0 GPa (0.435 106 psi).

Physics
2 answers:
torisob [31]3 years ago
6 0

Answer:

Maximum allowable surface crack length =  l =  0.000611m

Explanation:

Tensile stress = σ = 1.25 MPa = 1.25 × 10^6 N/m2

Surface energy of polystyrene = Es = 0.50 J/m2

Elastic Mudolus = E = 3 GPa = 3 × 10^9 N/m2

Maximum allowable surface crack length = l = ?

We know that:

       l = 2EEs/πσ^2 ....... (i)

By putttin the values in equation (i)

       l = (2)(0.50)(3 × 10^9)/(3.14)(1.25 × 10^6)^2

       l =  0.000611m

Hence, the Maximum allowable surface crack length will be 0.000611m.

Scorpion4ik [409]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

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tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0
3 years ago
While skateboarding at 19 km/h, Alana throws a tennis ball at 11 km/h to her friend Oliver. If Alana is the reference frame, the
kifflom [539]
Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.
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lora16 [44]

Answer:none

Explanation:

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3 years ago
A convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of
OLga [1]

Answer:

<h2>36cm from the surface</h2>

Explanation:

Equation of refraction of a lens is expression according to the formula given below;

\dfrac{n_2}{v} = \dfrac{n_1}{u}=  \dfrac{n_2-n_1}{R}

R is the radius of curvature of the convex refracting surface = 12cm

v is the image distance from the refracting surface

u  is the object distance from the refracting surface

n₁ and n₂ are the refractive indices of air and the medium respectively

Given parameters

R = 12 cm

u = \infty (since light incident is parallel to the axis)

n₁  = 1

n₂  = 1.5

Required

<em>focus point of the light that is incident and parallel to the central axis (v)</em>

<em />

Substituting this values into the given formula we will have;

\dfrac{1.5}{v} - \dfrac{1}{\infty}=  \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0=  \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}=  \dfrac{0.5}{12}\\\\

Cross multiply

1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm

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<h2>Answer: Radio waves  </h2><h2 />

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In addition, they are very useful to transport information, being important in telecommunications. They are used not only for conventional radio transmissions but also in mobile telephony and TV.  

It should be noted that since radio signals have large wavelengths, they can be diffracted around certain obstacles, such as hills and mountain ranges, preventing the signal from reaching its destination.  

Therefore, the correct option is C.

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