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konstantin123 [22]
3 years ago
14

PLS ANSWER ASAP!!

Physics
1 answer:
koban [17]3 years ago
4 0

Answer:

<em>The closest choice to the answer is 34.0 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion experienced by an object that is launched near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the Range or maximum horizontal distance traveled by the object is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The football is kicked at an angle of 30° with an initial velocity of 20 m/s. Replacing those values:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ)}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ)}{9.8}}=35.35\ m

The closest choice to the answer is 34.0 m

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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}

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