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Zielflug [23.3K]

3 years ago

5

Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If th

e distance is
68 cm when the weight is 2 kg, what is the distance when the weight is 5 kg?

1 answer:

zlopas [31]3 years ago

4 0

It would just be 2x68, in the sense that the original equation for Hooke's Law is F=kX. In your equation, w is k and X is D. the answer is 136cm for 2kg and 340cm for 5kg. Tell me if this is wrong please.

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2 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

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Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

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$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

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Where n is the nth term

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Substitute the above expression in equation 1

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$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

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By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

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Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

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$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

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So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

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Step-by-step explanation:

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