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Molodets [167]
4 years ago
11

The elastic potential energy is stored in a certain spring is 14 J when it is compressed 0.240m. the spring is now cut into thre

e equal pieces. once of the pieces is then compressed by 0.150m.
How much elastic potential energy is stored in the compressed piece?
Physics
2 answers:
ELEN [110]4 years ago
6 0

Answer:

5.47 J

Explanation:

From Hook's law,

Energy stored in a elastic material = 1/2ke²

E = 1/2ke²................. Equation 1

Where k = spring constant of the material, e = extension.

make k the subject of the equation

k = 2E/e²................... Equation 2

Given: E = 14 J, e = 0.24 m

Substitute into equation 2

k = 2(14)/0.24²

k = 28/0.0576

k = 486.11 N/m.

The elastic potential energy stored in the compressed piece is

Given: e = 0.15 m

E = 1/2(466.11)(0.15²)

E = 5.47 J

Yanka [14]4 years ago
4 0

Answer:

The elastic potential energy stored in the compressed piece is 5.47 J

Explanation:

Given;

elastic potential energy stored, E = 14 J

initial compression, x =  0.240m

The elastic potential energy stored in a spring is given as;

E = ¹/₂Kx²

where;

E is elastic potential energy

K is the spring constant

x is the compression

K = 2E / x²

K = (2*14) / (0.24²)

K = 486.11 N/m

When the spring is cut into three equal pieces, and one of pieces is then compressed by 0.150m, the energy stored in this piece is calculated as;

E = ¹/₂Kx²

E = ¹/₂(486.11)(0.15²)

E = 5.47 J

Therefore, the energy stored in one of the cut pieces compressed by 0.150m is 5.47 J

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A common practice in cooking is the addition of salt to boiling water (Kb = 0.52 °C kg/mole). One of the reasons for this might
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Answer:

640.59 g of NaCl.

Explanation:

Using Boiling point elevation,

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kb = 0.52 °C kg/mole

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5 0
4 years ago
An aluminum wire with a diameter of 0.100mm has a uniform electric field of 0.200V/m imposed along its entire length. The temper
Aleonysh [2.5K]
  1. The linear resistivity of this wire is equal to 3.15 × 10⁻⁸ Ωm.
  2. The current density in this wire is equal to 6.35 × 10⁶ A/m².
  3. The total current in a wire is equal to 0.0499 Amp.
  4. The drift speed of the conduction electrons is equal to 6.59 × 10⁻⁴ m/s.
  5. The potential difference between the ends of this wire is equal to 0.4 Volt.

<u>Given the following data:</u>

Diameter of aluminum wire = 0.100 mm.

Uniform electric field of aluminum wire = 0.200 V/m.

Temperature of aluminum wire = 50.0°C.

<u>Scientific data:</u>

Resistivity of aluminum, ρ = 2.82 × 10⁻⁸ Ωm

Temperature coefficient for aluminum, α = 3.9 × 10⁻³ °C⁻¹.

<h3>How to determine the resistivity?</h3>

Mathematically, the linear resistivity of a material can be calculated by using this formula:

ρ = ρ₀(1 + αΔT)

ρ = ρ₀(1 + α(T₂ - T₁)

ρ = 2.82 × 10⁻⁸ × [1 + 3.9 × 10⁻³(50 - 20)

Resistivity, ρ = 3.15 × 10⁻⁸ Ωm.

<h3>What is the current density in this wire?</h3>

Mathematically, the current density in a wire can be calculated by using this formula:

J = σE = E/ρ

J = 0.2/3.15 × 10⁻⁸

Current density, J = 6.35 × 10⁶ A/m².

<h3>What is the total current in this wire?</h3>

Mathematically, the total current in a wire can be calculated by using this formula:

I = JA = J(πr²)

I = 6.35 × 10⁶ × (3.142 × 0.00005²)

Total current, I = 0.0499 Amp.

<h3>What is the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = I/nqA

V = (0.0499 × 0.027)/(6.023 × 10²³ × 27000 × 1.602 × 10⁻¹⁹ × (3.142 × 0.00005²)

Drift speed, V = 6.59 × 10⁻⁴ m/s.

For the the potential difference, we have:

Mathematically, the potential difference between the ends of a wire can be calculated by using this formula:

ΔV = El

ΔV = 0.2 × 2

ΔV = 0.4 Volt.

Read more on drift speed here: brainly.com/question/15219891

#SPJ4

Complete Question:

An aluminum wire with a diameter of 0.100 mm has a uniform electric field of 0.200 V/m imposed along its entire length. The temperature of the wire is 50.0°C. Assume one free electron per atom.

(a) Determine the resistivity.

(b) What is the current density in the wire?

(c) What is the total current in the wire?

(d) What is the drift speed of the conduction electrons?

(e) What potential difference must exist between the ends of a 2.00-m length of the wire to produce the stated electric field?

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Use equation for accelaration : a=(V1-V0)/t

a=(0m/s-27.7m/s)/10s

a=-27.7s/10s

a=2.77m/s*s

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