PLEASE HELP ASAP!!<br><br> Which of the following diagrams represents a complete series circuit?

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinal and transverse modul

gregori [183]

Answer:

It is not possible to produce continued and oriented fiber.

Explanation:

To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:

$E = (1-V_f)E_m+V_fE_f$

Where,

E is the longitudinal modulus of Elasticity

$V_f$ is the fiber volume ratio

$E_m$ is the elastic modulus of the matrix

$E_f$ is the elastic modulus of the fibers

We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is

$E_f = 131Gpa$

Moreover from the statement,

$E = 35Gpa$

$E_m = 3.4Gpa$

Replacing in the previous equation,

$35 = 3.4 (1-V_f)+131V_f$

$V_f = 0.25 \rightarrow longitudinal$

To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,

$E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}$

Our values are given in this case as:

$E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa$

Replacing,

$5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}$

$V_f = 0.351 \rightarrow transversal$

From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.

5 0

3 years ago

approximately what force,FM must the extensor muscle in the upper arm exert on the lower arm to hold a 7.6kg shot put? assume th

MAXImum [283]

Lets do the sum of the forces about the elbow joint.

Fm = Force of Muscle; Fe = Force Elbow; Fb = Force Ball

Sum Force about Joint = (-2.5)Fm + 12.5Fe + 30Fb = 0

(-2.5)Fm + 12.5(2.8) + 30(6.9) = 0

Fm = 96.8kg

Fm = 96.8 * 9.8 = 948.6N

Do you understand why the -2.5 is negative?
<span> Because I put the origin at the joint. So when you go left it is negative and when you go right it is positive. </span>

7 0

3 years ago

Which describes how the same force affects a small mass and a large mass

goldenfox [79]

You shall use the equation for force given by the second Law of Newton, this is F = m*a, where F is the net force that acts over the object, m is the mass of the object and a is the acceleration that the object will acquire. From that equation you can find a = F/m, which means that a is direct proportional to F and invsersely related to m. So, small masses accelerate faster than large masses, and <span>the answer is the option B. the small mass accelerates faster.</span>

5 0

3 years ago

Two forces of magnitudes 4.0 Newtons and 3.0 Newtons pull on a box. The forces make an angle of 40° with each other. What is the

leonid [27]

Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

$F_{1x} = 4.0 N\\F_{1y} = 0$

$F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N$

So the resultant has components

$F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N$

So the magnitude of the resultant is

$F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N$

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.

3 0

2 years ago

What is the acceleration of a 10kg mass pushed by a 5N force?

GenaCL600 [577]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

$a = \frac{5}{10} = \frac{1}{2} \\ = 0.5$

We have the final answer as

Hope this helps you

5 0

2 years ago

PLEASE HELP ASAP!! Which of the following diagrams represents a complete series circuit?