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2 years ago

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La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el

peso del auto y el cuadrado de la velocidad del mismo. Supongamos que 400 libras de fuerza evitan que un auto que pesa 1600 libras derrape en una curva cuyo radio mide 800 si viaja a 50mph. ¿Cuánta fuerza evitaría que el mismo auto derrapara en una curva cuyo radio mide 600 si viaja a 60mph ?

1 answer:

Vika [28.1K]2 years ago

7 0

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

400 = k × 5000

k = 400/5000

k = 2/25

Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

Por lo tanto, la cantidad de fuerza que evitaría que el mismo automóvil patine en una curva con un radio de 600 si viaja a 60 mph es de 768 libras.

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Answer:

First Case:

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Second Case:

$\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}$

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

$6p+2, 4p^2-10, \text{ and } 4p+3$

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

$4p^2-10=(6p+2)+d$

And, likewise, for the third term:

$4p+3=(6p+2)+2d$

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

$d=4p^2-6p-12$

And for the second equation:

$2d=-2p+1$

To avoid fractions, let's multiply the first equation by 2. Hence:

$2d=8p^2-12p-24$

Therefore:

$8p^2-12p-24=-2p+1$

Simplifying yields:

$8p^2-10p-25=0$

Solve for <em>p</em>. We can factor:

$8p^2+10p-20p-25=0$

Factor:

$2p(4p+5)-5(4p+5)=0$

Grouping:

$(2p-5)(4p+5)=0$

Zero Product Property:

$\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}$

Then, we can use the second equation to solve for <em>d</em>. So:

$2d_1=-2p_1+1$

Substituting:

$\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}$

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

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So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

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Anon25 [30]

Answer:

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As the rate of fill is constant,

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From the table,

Volume of water left after $t=2$ minutes is 184 liters.

So, $x-R(2)=184\\x-2R=184$ -------- 1

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So, $x-R(12)=4\\x-12R=4$ ------------2

Multiply equation 1 by 6 and equation 2 by -1, we get

$6x-12R=1104\\-x+12R=-4$

Now, we add the above equations, we get

$(6x-x)+(12R-12R)=1104-4\\5x=1100\\x=\frac{1100}{5}=220$

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Answer:

3 1/3 cups of flour

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