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andrew-mc [135]
3 years ago
13

You also sell £350 worth of £25 maths starter kits. You think you sold 14. Select the correct inverse calculation below that wou

ld allow you to check your answer.
Mathematics
1 answer:
Contact [7]3 years ago
6 0
$350×25x=14 is the equation for this question
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Find the GCF for 15az and 25az
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En la tienda de mascotas "Animalo-T", se desea elevar un elefante de 2,900 kg utilizando una elevadora hidráulica de plato grand
Korvikt [17]

Answer:

Se requiere una fuerza de 854.473 newtons sobre el émbolo pequeño.

Step-by-step explanation:

Por el Principio de Pascal se conoce que el esfuerzo experimentado por el elefante es igual a la presión ejercida por el plato pequeño. Es decir:

\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} (1)

Donde:

F_{1} - Fuerza experimentada por el elefante, medida en newtons.

F_{2} - Fuerza aplicada sobre el plato pequeño, medida en newtons.

A_{1} - Área del plato grande, medida en metros cuadrados.

A_{2} - Área del plato pequeño, medida en metros cuadrados.

La fuerza aplicada sobre el plato pequeño es:

F_{2} = \left(\frac{A_{2}}{A_{1}} \right)\cdot F_{1}

La fuerza experimentada por el elefante es su propio peso. Por otra parte, el área del plato es directamente proporcional al cuadrado de su diámetro. Es decir:

F_{2} = \left(\frac{D_{2}}{D_{1}} \right)^{2}\cdot m\cdot g (2)

Donde:

D_{1} - Diámetro del plato grande, medido en centímetros.

D_{2} - Diámetro del plato pequeño, medido en centímetros.

m - Masa del elefante, medida en kilogramos.

g - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que D_{1} = 0.75\,m, D_{2} = 0.13\,m, m = 2900\,kg y g = 9.807\,\frac{m}{s^{2}}, entonces la fuerza a aplicar al émbolo pequeño es:

F_{2} = \left(\frac{0.13\,m}{0.75\,m} \right)^{2}\cdot (2900\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F_{2} = 854.473\,N

Se requiere una fuerza de 854.473 newtons sobre el émbolo pequeño.

3 0
3 years ago
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