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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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Write a complete scientific explanation to account for why the ball that was moving faster caused more flour to spread out.​
Harlamova29_29 [7]

Answer:

It is due to the large impulse is imparted on the flour.

Explanation:

A ball is moving faster.

When a ball is moving faster strikes to the flour, the change in momentum is large and thus the impulse imparted on the flour is large.

Impulse = change in momentum

So, as the flour experiences large impulse and large momentum so that the flour spreads out.

If the change in momentum is large so the flour spreads out is more.  

8 0
3 years ago
6. What are the four chambers of the heart? A. atrium, artery, ventricle, vein B. artery, blood vessel, capillary, vein C. blood
neonofarm [45]

Answer:

The answer is D. right atrium, left atrium, right ventricle, left ventricle

Explanation:

The right atrium receives oxygen-poor blood from the body and pumps it to the right ventricle.The right ventricle pumps the oxygen-poor blood to the lungs.The left atrium receives oxygen-rich blood from the lungs and pumps it to the left ventricle.The left ventricle pumps the oxygen-rich blood to the body.

3 0
3 years ago
List three measurements with different units this are equal to 5 meters
olganol [36]
5kg
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Hope this helped good luck to you
6 0
3 years ago
The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0
3 years ago
An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.
amm1812

Answer:

a) T ’= 0.999 s ,  b)  t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

        w = √g / L

The angular velocity, frequency and period are related

        w = 2π f = 2π / T

        2π / T = √ g / L

        T = 2π √ L / g

        L = T² g / 4π²

        L = 1² 9.8 / 4π²

        L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

       ΔL = α L ΔT

       ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

       ΔL = 142.8 10⁻⁶ m

       Lf - L = -142. 8 10⁻⁶

       Lf = 142.8 10⁻⁶ + 0.248

       Lf = 0.2479 m

Let's calculate new period

      T ’= 2π √ L / g

      T ’= 2π √ (0.2479 / 9.8)

      T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

       t = 3600 0.999

       t = 3596.4 s

Therefore the clock is delayed almost 4 s

6 0
4 years ago
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