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3 years ago

What happens to the rock structure during each of type of change?

1 answer:

3 0

Previous rocks melt and collide and to form igneous rocks.
Igneous rocks disintegrate due to weather disruptions and get carried away by water, where they form sedimentary rock strata by lithification.
Igneous and sedimentary change by heat and pressure to form metamorphic rocks.
Metamorphic rocks melt and become igneous rocks.

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Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago

a 100 g cart initially moving at 0.5 m/s collides elastically from a stationary 180 g cart. a) using the equation in the theory,

cupoosta [38]

A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.

Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.

Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.

For an elastic collision, we use the formula,

m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f

For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.

Final velocity = 0.5/2

=0.25 m/s.

To know more about elastic collision, refer

brainly.com/question/7694106

#SPJ4

7 0

1 year ago

In a grassland ecosystem, praying mantises catch and eat grasshoppers. However, there has been a significant decrease in the pra

Alekssandra [29.7K]

Answer:

The population will increase because grasshoppers are the prey.

Explanation:

If there has been a significant decrease in the praying mantis population in recent years. This most likely affect the grasshopper population over time by the population will increase because grasshoppers are the prey. If there is less praying mantis then grasshoppers won't be eaten as much so the population of the grasshoppers will thrive

7 0

3 years ago

Read 2 more answers

A car of weight 300N moved through a distance of 10m when pushed by 3 students

Sati [7]

The work done by the three students is 3,000 J.

The energy transferred in the process is 3,000 J.

Work done is the product of force and distance moved by the object.

W = Fd

The work done by the three students is calculated as follows;

W = 300 x 10

W = 3,000 J

<h3>What is energy transfer?</h3>

This is means by which energy is converted from one form to another.

The energy transferred in the process is determined by work energy theorem.

E = W

E = 3,000 J

Learn more about work-energy theorem here: brainly.com/question/22236101

3 0

2 years ago

A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass

natka813 [3]

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet = 3.56×10²⁶ Kg.

5 0

3 years ago