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2 years ago

15

Demarcus launches a small weight into the air. The weight takes 6.4 seconds to reach the ground again. At what time did the weig

ht have a vertical velocity of 0 meters per second?

1 answer:

Monica [59]2 years ago

5 0

The answer is A.
During the process, the only force imposed on the small weight is gravity.
The time from launching to the highest is the same as the time from the highest to the ground. And the vertical velocity at 0 m/s is at the highest position.

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Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3

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Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

$k_{1}=20\ N/m$

$k_{2}=30\ N/m$

$k_{3}=15\ N/m$

$k_{4}=20\ N/m$

$k_{5}=35\ N/m$

According to figure,

$k_{2}$ and $k_{3}$ is in series

We need to calculate the equivalent

Using formula for series

$\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}$

$k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}$

Put the value into the formula

$k=\dfrac{30\times15}{30+15}$

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k and $k_{4}$ is in parallel

We need to calculate the k'

Using formula for parallel

$k'=k+k_{4}$

Put the value into the formula

$k'=10+20$

$k'=30\ N/m$

$k_{1}$ ,k' and $k_{5}$ is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

$k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}$

Put the value into the formula

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Evgesh-ka [11]

Answer:

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