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AlexFokin [52]
4 years ago
13

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

of elasticity of 35 GPa (5×106psi) and 5.17 GPa (7.5×105psi), respectively? Why or why not?Assume that the elastic modulus of the epoxy is 3.4 GPa (4.93×105psi).Also, the value ofEfor aramid fibers is 131GPa.
Physics
1 answer:
Masja [62]4 years ago
5 0

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

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Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy =\frac{1}{2} mv^2, where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0
3 years ago
What are the four most common gases in dry air?
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Nitrogen
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7 0
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A 1.45 kg falcon catches a 0.515 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is
True [87]

Answer:

Their velocity after the impact is 20.85 m/s.                            

Explanation:

Given that,

Mass of falcon, m_1=1.45\ kg

Mass of dove, m_2=0.515\ kg

Initial speed of the falcon, u_1=26.5\ m/s

Initial speed of the dove, u_2=4.95\ m/s

We need to find the final velocity after the impact. When the falcon catches the dove, it will becomes the case of inelastic collision. The conservation of momentum will be :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{1.45\times 26.5+0.515\times 4.95}{(1.45+0.515)}\\\\V=20.85\ m/s

So, their velocity after the impact is 20.85 m/s.                          

8 0
3 years ago
I need help pleaseee
Ludmilka [50]

Answer:

8) 709.8875 J

9) The object is at 7.24375 m from the ground

10) Kinetic energy increases as the object falls.

Explanation:

We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:

h(t) = hi - (g/2) t^2

hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):

h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m

This is the position of the 10 kg object after 0.75 seconds (answer for part 9)

Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:

U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)

Part 10)

the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.

3 0
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f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If
Luden [163]

Answer:

The angular velocity is   w= \sqrt[4]{\frac{a_{max}^2}{r^2}  - \alpha ^2}      

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

          The Radial acceleration which is mathematically represented as

                              a_r = \frac{v^2}{r}  = w^2r

And the Tangential  acceleration which is mathematically represented as

                                a_r = \alpha r

  The net acceleration is evaluated as

                      a = \sqrt{a_r^2 + a_t^2}

       

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

                 

So at maximum angular acceleration we a have

             a_{max} = \sqrt{a_r^2 + a_t^2}

                     a_{max}^2 = a_r^2 + a_t^2

                    a_{max}^2 = (w^2r)^2 + (\alpha r)^2

                 a_{max}^2 =  r^2 w^4 + r^2 \alpha ^2

                  a_{max}^2 = r^2 (w^4 + \alpha^2 )

                w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}

                         w^4 = \frac{a_{max}^2}{r^2}  - \alpha ^2

                         w= \sqrt[4]{\frac{a_{max}^2}{r^2}  - \alpha ^2}        

                     

3 0
3 years ago
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