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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.

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Karo-lina-s [1.5K]

<u>Correct Question:</u>

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

<u>Answer:</u>

The total distance covered by Charlie is 8 km in 1 hour.

<u>Explanation:</u>

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$t=1 hour

As we know the formula to evaluate the total distance d when the average velocity and time is given;

v=\frac{d}{t}

d=v \times t

d=8 \times 1

d=8 k m

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

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How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?
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To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

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what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

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Solving the equation,we find

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katen-ka-za [31]

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3 years ago
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Use Newton’s Universal Law of Gravitation to calculate the magnitude of the gravitational force between a 200 kg refrigerator an
expeople1 [14]

Answer:

3.735×10⁻⁶ N

Explanation:

From newton' s law of universal gravitation,

F = Gmm'/r² .............................. Equation 1

Where F = Gravitational force between the person and the refrigerator, m = mass of the person, m' = mass of the refrigerator, r = distance between the person and the refrigerator. G = gravitational universal constant.

Given: m = 70 kg, m' = 200 kg, r = 0.5 m

Constant: G = 6.67×10⁻¹¹ Nm²/kg².

F = (6.67×10⁻¹¹×70×200)/0.5²

F = 93380×10⁻¹¹/0.25

F = 373520×10⁻¹¹

F = 3.735×10⁻⁶ N

Hence the force between the person and the refrigerator =  3.735×10⁻⁶ N

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