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3 years ago

How to describe the motion, and velocity of a moving man

Physics

1 answer:

Kazeer [188]3 years ago

7 0

Answer:

You can describe the motion of an object by its position, speed, direction, and acceleration. An object is moving if its position relative to a fixed point is changing. Even things that appear to be at rest move.

Explanation:

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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

What happened to the kinetic energy when the velocy of a moving body is doubled?

dem82 [27]

Answer:

If velocity is doubled, Kinetic Energy increases by 4 times. Kinetic energy of a body is the energy possessed by it, by virtue of its motion, i.e. if the body is moving it will always have kinetic energy.Nov 1, 2018

Explanation:

8 0

3 years ago

Read 2 more answers

What does the slope on a velocity-time graph measure?

Lostsunrise [7]

Answer:The slope in the velocity-time graph represents the acceleration. The slope is defined as the ratio of change in y-axis to change in the x-axis. The slope is represented by the letter m and following is the general formula used for determining the slope:

Explanation:

6 0

3 years ago

Read 2 more answers

Un cuerpo se mueve con MRU a una rapidez de 12 m/s y recorre 240 m; luego, disminuye su ra¬pidez uniformemente a razón de 2 m/s2

Annette [7]

Answer:

29 seconds

Explanation:

First we have a constant speed of 12 m/s and the distance of 240 m, so to find the time we can use the formula:

distance = speed * time

240 = 12 * time1

time1 = 20 seconds

Then, the speed decreases at 2 m/s2 until it reaches 2 m/s. So to find this time, we use this formula:

Final speed = inicial speed + acceleration * time

2 = 12 - 2 * time2

2*time2 = 10

time2 = 5 seconds.

Then, the speed increases from 2 m/s to 22 m/s with an acceleration of 5 m/s2, so we have:

Final speed = inicial speed + acceleration * time

22 = 2 + 5 * time3

5*time3= 20

time3 = 4 seconds

The total time is:

Total time = time1 + time2 + time3 = 20 + 5 + 4 = 29 seconds

7 0

3 years ago

A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the s

tamaranim1 [39]

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

$vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j$

(a) The velocity of boat with respect to ground is

$\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s$

(b) The direction is given by

$tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o$

7 0

3 years ago