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Lesechka [4]
3 years ago
10

how much energy is stored in a 7.0 cm spring that has a spring constant of 500 N/m when it is stretched 3.0 cm​

Physics
1 answer:
zaharov [31]3 years ago
8 0
<h3>Answer:</h3>

12250 J

<h3>Explanation:</h3>

<em>elastic potential energy = 0.5 × spring constant × (extension)2</em>

spring constant = 500

extension =  7cm

7^2=49

0.5 × 500 × 49 = 12250 J or 12.25kJ

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AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × 10^{-6} m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top  t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = 10^{5} Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

so

pressure at bottom P2 = 10^{5} + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × 10^{5}  Pa

so from gas law

\frac{P1*V1}{t1} = \frac{P2*V2}{t2}

here p is pressure and v is volume and t is temperature

so put here value and find v1

\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0
3 years ago
Calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
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Answer:

58.8J

Explanation:

Given parameters;

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Unknown:

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Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

  Potential energy = mass x acceleration due to gravity x height

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8 0
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A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r
katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

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Where σ is stress given by the formula Force(F)/Area(A)

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Thus;

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So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

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