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qwelly [4]
3 years ago
6

The diagonal of a TV is 42 inches. The base is 38 inches wide. How tall is the Tv? Round your answer to the nearest tenth. Put y

our answer in the box. EX: 10
Mathematics
2 answers:
Yuki888 [10]3 years ago
4 0

Answer:20.6 inches

Step-by-step explanation:

The diagonal measure of a rectangular TV screen is 42 inches. The width of the screen is 36.6 inches.

The diagonal , width and height forms a right angle triangle

So we apply pythagorean theorem

Diagonal is the hypotenuse

LEt the height of the screen is 'h'

c^2= a^2 + b^2

42^2= 36.6^2 + h^2

Subtract 36.6^2 from both sides

h^2 = 424.44

Take square root on both sides

h=20.60194

Round to nearest tenth

Height of the screen = 20.6 inches

iVinArrow [24]3 years ago
4 0

Answer:

1,600 inches

Step-by-step explanation:

You need to take the height of the tv and the base of the tv. So you need to take 42x38= 1,596. Now you need to round to the nearest tenth. So take 1,596 and the nearest tenth would be 1,600. Because 1,596 is closer to 1,596 then 1,590.

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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

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