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love history [14]

3 years ago

15

A 1.35 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.8 n/m. the spring is initially unstr

etched. a constant 23.8 n horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s 2 . 23.8 n 1.35 kg 19.8 n/m find the speed of the block after it has moved 0.236 m from equilibrium if the surface between block and tabletop is frictionless. answer in units of m/s.

1 answer:

Ket [755]3 years ago

7 0

Horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s

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A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6

liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

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Therefore,

90 revolutions in 1 minute

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Hatshy [7]

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

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$k=10\ N/m$

k and $k_{4}$ is in parallel

We need to calculate the k'

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$k'=k+k_{4}$

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

$k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}$

$k_{eq}=8.93\ N/m$

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

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Natasha_Volkova [10]

Answer:

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