Problem 2
Plot point L anywhere that isn't on segment JK. Draw a line through point L. I find it helps to make the lines parallel.
Next, use a compass to measure the width of segment JK. Keeping this same width, transfer the nonpencil end of the compass to point L. Draw an arc that crosses the line through L.
Mark this intersection point M. Lastly, use a pen or marker to form segment LM and erase everything else of that line.
======================================================
Problem 3
The ideas of the previous problem will be used here. We copied segment JK to form congruent segment LM. So JK = LM.
The same steps will be used to form segment GN where GN = EF. In other words, segment GN is a perfect copy of segment EF.
If you repeat these steps again, you'll get another segment of the same length. This segment goes from point N to point H. So NH = GN = EF
Then we can say,
GH = GN + NH
GH = EF + EF
GH = 2*EF
#8 Z=14
#9 D=98
#10 F=622
# 11 A=27
I hope this helps!!
If George completed 3/5 of work in 9 days, it means that he needs 3 days to finish 1/5th of the work.
The remaining part is 2/5 because 1-3/5=2/5. 2/5 is also 6/15
From this, how much did George do? in the first 3 days he did 1/5th and then one more day was left, during which he did 1/5/3=1/15th.
So he did in total 1/15+1/5=1/15+3/15=4/15.
this means that Paul did the rest of 6/15, so Paul did 6/15-4/15=2/15.
So we know that he does 2/15 in 4 days, which means that every two days he can do 1/15 of the work
so he would need 15 times 2 days to finish the work - 30 days!
El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.
<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>
La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.
Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:
σ = W / (π · D² / 4)
Donde:
- W - Peso de la caja, en newtons.
- D - Diámetro del área transversal de la caja, en metros.
Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:
σ = 25 N / [π · (0.02 m)² / 4]
σ ≈ 79577.472 Pa
<h3>Observación</h3>
La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.
Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145
#SPJ1
76.
If the number after the decimal is below five, you round down. If the number is above five, you round up. 76 is your answer.
