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Mila [183]
2 years ago
12

An engineer uses aluminum to build an airplane rather than composite materials that are lighter and stronger. He does this becau

se he is concerned about the composite materials and how suddenly they can fail. This is an example of:
Physics
2 answers:
NARA [144]2 years ago
7 0

choosing a material that will show warning before it fails

just checking all credits go to the other person

AleksandrR [38]2 years ago
4 0

Answer:

choosing a material that will show warning before it fails

Explanation:

According to my research on different architectural engineering techniques, I can say that based on the information provided within the question this is an example of choosing a material that will show warning before it fails. By choosing aluminum he can detect certain failures a long time before it actually happens since aluminum shows signs of wear and tear and doesn't just break immediately.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i
laiz [17]
L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ =  the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer:  15 mm
4 0
2 years ago
A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction bctwecn thc box and the f
Hitman42 [59]

Answer:

(D) 96 kg-m/s

Explanation:

Let's start off by first calculating the normal force between the box and the floor.

This will be:

Normal Force = 12 * 9.81 = 117.72 N

We can now use the friction equation to find the frictional force on the box when it is moving:

Frictional force = Coefficient of friction * Normal Force

Frictional force = 0.4 * 117.72 = 47.09 N

Finally, since we have the force on the box, we can find the acceleration:

F = Mass * Acceleration

47.09 = 12 * Acceleration

Acceleration = 3.92 m/s^2

Final speed after 2 seconds:

V=U+a*t

V = 4 +(-3.92)*(2)

V= -3.84 m/s

Since we know the initial and final speeds, we can calculate the change in momentum:

Change in momentum = Final Momentum - Initial Momentum

Change in momentum = 3.84*12-(-4)*12

Change in momentum = 94.08 kg*m/s

Thus we can see that option (D) is the closest answer.

6 0
2 years ago
10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
2 years ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
2 years ago
Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su
telo118 [61]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

4 0
2 years ago
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