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Bond [772]
3 years ago
14

A block slides on a frictionless, horizontal surface with a speed of 1.32 m/s. The block encounters an unstretched spring and co

mpresses it. If it takes the block 0.4 s to come to rest, what would be hte period of oscillation if the block stays stuck to the spring? If the spring extends back the other direction, the block eventually loses contatct with the psring and slides away, what is the velocity of the block when it loses contact with the spring?
Physics
1 answer:
Rus_ich [418]3 years ago
3 0

Answer:

Explanation:

The given time is 1 / 4 of the time period

So Time period  of oscillation.

= 4 x .4 =1.6 s

When the block reaches back its original position when it came in contact with the spring for the first time , the block and the spring will have maximum

velocity. After that spring starts unstretching , reducing its speed , so block loses contact as its velocity is not reduced .

So required velocity is the maximum velocity of the block while remaining in contact with the spring.

v ( max ) = w A = 1.32  m /s.

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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

4 0
3 years ago
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6 0
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Answer:

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Explanation:

7 0
3 years ago
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Answer:

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5 0
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Explanation:

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