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3 years ago

10

A vertically hung spring has a spring constant of 150. newtons per meter. A

1 answer:

BARSIC [14]3 years ago

7 0

Based on Hooke's law, the force a spring exerts is negative based on the elongation length. Therefore, since a 2.00 kilogram weight exerts 20 Newtons (gravitation acceleration constant = 10), than the spring must pull back with -20 Newtons. This means that -20 = -kx, where k = spring constant and x = elongation length. k = 150, so that means that x = 20/150, or 2/15 meters.

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What two variables is acceleration dependent on?

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4 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr

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Answer:

48.4293354946 N

Yes

Explanation:

d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

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3 years ago

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3 years ago

Read 2 more answers

If x =(20-4t^2) find average velocity between t=0and t=2sec

lapo4ka [179]

Answer:

The average velocity is 8 unit per sec

Explanation:

Given as :

The Distance x = 20 - 4t² unit

Change in time Δt = ( 2 - 0 ) s = 2 s

Let the velocity = V unit/s

∴ V = $\frac{\partial (20 - 4t^{2})}{\partial x}$

Or, V = - 8t unit/s

Now velocity at t = 0

V1 = - 8 × 0 = 0 unit/s

And velocity at t = 2 sec

V2 = - 8 × 2 = - 16

So, Average velocity = $\frac{v2 - v1}{2}$ = $\frac{- 16 - 0}{2}$ = -8

Or, $\begin{vmatrix}V\end{vmatrix}$ = 8 unit/sec

Hence The average velocity is 8 unit per sec Answer

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3 years ago