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docker41 [41]
3 years ago
6

The range of the quedractic function is {y|y

Mathematics
2 answers:
solmaris [256]3 years ago
8 0

Answer: The question is incomplete, Hope I could've helped.

Evgesh-ka [11]3 years ago
6 0

well I have no idea what that means but I hope you find your answer

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Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?
Naddik [55]

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

5 0
1 year ago
If a bullentin board is 6 ft wide and 4 ft tall. what are the deminsions inside
SOVA2 [1]
If you are looking for the area the answer would be 24 ft. tall because 6*4 is 24
6 0
4 years ago
Can someone help me with this ASAP please I’m being timed !
Paladinen [302]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
What is the circumference of a 9 m circle?
zhenek [66]
From what I know its 56.52m
5 0
3 years ago
50 points! I understand A. and B. but I would really appreciate help with C.
FromTheMoon [43]

Answer:

51.72\text{ cells per hour}

Step-by-step explanation:

So, the function, P(t), represents the number of cells after t hours.

This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.

C)

So, we are given that the quadratic curve of the trend is the function:

P(t)=6.10t^2-9.28t+16.43

To find the <em>instanteous</em> rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:

\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]

Expand:

P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]

Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:

P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]

Differentiate. Use the power rule:

P'(t)=6.10(2t)-9.28(1)

Simplify:

P'(t)=12.20t-9.28

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

P'(5)=12.20(5)-9.28

Multiply:

P'(5)=61-9.28

Subtract:

P'(5)=51.72

This tells us that at <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.

And we're done!

7 0
3 years ago
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