1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
madreJ [45]
3 years ago
9

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

gth of a material are tensile strength and shear strength. The tensile strength of a material is the amount of force needed to pull the material apart perpendicular to a cross section, such as pulling on a rope. The shear strength is the amount of force needed to rupture the material when a force is applied parallel to the cross section, such as breaking a pencil overhanging a table edge. Both tensile strength and shear strength have units of pressure, because they represent the force that must be applied per unit area in order to break the material. The importance of knowing the tensile strength and shear strength of a material is evidenced by the number of machines and companies specializing in determining these properties. Many types of materials and objects are tested, from fabric and metal to nuts and bolts. It is always comforting to know that the tensile strength of the rope holding you above the ground can support your weight. Shear strength needs to be found for bridges, the floors of multilevel buildings, and the joints where the wings join the body of an airplane. Without the experimental determination of these properties, trial and error would be the only other method of determining these values, a process that would be dangerous to life and limb.
QUESTION A) Given that the tensile strength of aluminum foil is 311 megapascals, its thickness is approximately 15.0 micrometers, and a roll of household aluminum foil is 30.0 centimeters wide, how much force F is needed to pull off a sheet to use?

Assume that the force is applied equally along the whole width of the foil, and that you are trying to pull perpendicular to the cross section.

QUESTION B)

As stated in the introduction, shear strength is another measure of the strength of a material. A shear force is a force that acts parallel to the plane in the material that breaks. A good example of a shear is that of a martial arts expert breaking boards or bricks with her hands. Other applications in which shear forces and shear strength need to be known are geology, for studying earthquakes and landslides; fluid dynamics; and structural engineering.

Aluminum has a shear strength of 210 megapascals. When you bend aluminum foil around an edge (i.e., the edge of the box) and pull, you are effectively applying a shear force along the bent edge of the foil. If a roll of household aluminum foil is 30.0 centimeters wide and its thickness is approximately 15.0 micrometers, how much shear force is needed to pull off a sheet?

Assume that the force is applied equally along the whole width of the foil.
Physics
1 answer:
katrin2010 [14]3 years ago
6 0

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

You might be interested in
Magnetism is a
san4es73 [151]
Your answer would be a, it is a  force.
6 0
3 years ago
Read 2 more answers
A positively charged particle is moving toward the South in a region of space where there is a magnetic field directed down (tow
Alexxandr [17]

Answer:

F = Q V X B

V = south

B = down

V X B = east

8 0
2 years ago
A hypothesis is
Feliz [49]
The answer is C, an educated guess
3 0
3 years ago
as the video shows, the star begins its life from a clump of gas that heats up as it contracts. where does the energy that heats
MAXImum [283]

When water vapor condenses during the formation of clouds, energy is released into the atmosphere through the process of latent heating.

Huge amounts of solar energy are absorbed during the evaporation process. There is no thermal equilibrium in protostars. They emit radiation that causes them to lose energy, but their interior temperatures aren't hot enough to start nuclear fusion to replenish the energy. They consequently tighten and warm up. Fusion can replace the energy they are losing once they are heated enough on the inside (about 107 K). The gas atoms start to collide more and get closer together as the cloud gets smaller, which warms them up. The cloud keeps shrinking and heating up until the force of gravity is balanced by the pressure from the heat.

Learn more about energy here-

brainly.com/question/1932868

#SPJ4

5 0
2 years ago
A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = \frac{V}{R}

where R is the resistance

R = \frac{pL}{r^{2} \pi }

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = \frac{\pi(r)^{2}V  }{pL}

q = \frac{\pi(r^{2} )Vt }{pL}

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0
3 years ago
Other questions:
  • Most of the funding for research comes from the federal government or ? And is provided to Principal Investigators (PIs) through
    10·1 answer
  • Fossils are dated using radioactive...<br><br> a. hydrogen <br> b. helium<br> c. carbon
    9·2 answers
  • Can someone help me with this really quick please
    8·1 answer
  • A cat has a mass of 3 kg and runs at a speed of 6 m/s. how much kinetic energy does the cat have?
    12·1 answer
  • At what point in the swing of the pendulum is the potential energy completely converted into kinetic energy?
    13·1 answer
  • A friction-less pulley in the shape of a solid cylinder of mass 2.90 kg and radius 30.0 cm is used to draw water from a well. A
    7·1 answer
  • What is characteristic of an opaque object.
    8·2 answers
  • How long will it take for a rock to fall 50 meters
    12·1 answer
  • An object with a mass of 6.5 kg accelerates 12.3 m/s2 when an unknown force is applied to it. What is the amount of the force?
    7·1 answer
  • Conductors of electricity *<br> A.Nonmetal<br> B-Metalloid<br> C-Metal
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!