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laiz [17]
3 years ago
15

Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2

column table with 4 rows. The first column is labeled spring with entries W, X, Y, Z. The second column is labeled spring constant in newtons per meter with entries 24, 35, 22, 15. Which lists the springs based on the amount of elastic potential energy, from greatest to least? X, Y, W, Z X, W, Y, Z Z, W, Y, X Z, Y, W, X
Chemistry
1 answer:
svp [43]3 years ago
9 0

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S),         Spring Constant (N/m)

      W,                   24

      X,                    35

      Y,                    22

      Z,                    15

The equation for elastic potential energy, E_e, is E_e = 0.5 \times k \times x^2

The above equation can also be written as E_e =\dfrac{1}{2}  \times k \times x^2

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, E_e, of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, E_e, therefore the correct order is as follows;

X > W > Y > Z

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Answer:

S(metal) = 0.66J/g°C

Explanation:

We can find specific heat of a material, S, using the equation:

q = m*S*ΔT

<em>Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.</em>

The heat given by the metal is equal to the heat that water absorbs, that is:

m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)

<em>Where:</em>

m(Metal) = 76.0g

S(metal) = ?

ΔT(Metal) = 96.0°C-31.0°C = 65.0°C

m(Water) = 120.0g

S(water) = 4.184J/g°C

ΔT(water) = 31.0°C-24.5°C = 6.5°C

Replacing:

76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C

S(metal) = 0.66J/g°C

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The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.

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How many grams of butane were in 1. 000 atm of gas at room temperature?
dimaraw [331]

The mass in grams of butane at standard room temperature is 53.21 grams.

<h3>How can we determine the mass of an organic substance at room temperature?</h3>

The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:

PV = nRT

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1 × 22.4 L = n × (0.0821 atm*L/mol*K×  298 K)

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mass of butane = 53.21 grams

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The by-product of the chlorination of an alkane is​
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The by-product of the chlorination of an alkane is​  <u>HCl</u>

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