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Softa [21]
3 years ago
6

Which of the following statements is TRUE concerning forces?

Physics
1 answer:
Gnesinka [82]3 years ago
3 0
The answer is d I took the test
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Give two examples of Newton's second law of motion
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1. A 1.32 x 104 meter steel railroad track with a coefficient of linear expansion of 12 x 10-6 per degree Celsius changes temper
GarryVolchara [31]

Answer:

1) 0.1584 m

2) To allow for expansion without derailment

3) 0.101376 m

4) 213.675 °C

5) 266.67 m

6) 8.33 × 10⁻⁶ /°C

7) The alloy meets the requirement

8) 1.95 × 10⁻³ /°C

9) 32.095 m

10) -12157.72°C

Explanation:

1) Equation for the coefficient of linear expansion = \frac{\Delta L}{L} = \alpha _L \Delta T

Where:

ΔL = Change in length = Required

L = Initial length = 1.32 × 10⁴ m

\alpha _L = Coefficient of linear expansion of steel = 12 × 10⁻⁶ /°C

ΔT = Change in temperature = 37°C - 27°C = 10°C

Plugging the values in the equation for the temperature expansion of steel, we have m;

ΔL = L × \alpha _L ×ΔT = 1.32 × 10⁴ × 12 × 10⁻⁶ × 10  = 0.1584 m

2. Here we have that by segmenting railroad tracks into short pieces, the expansion of the metal tracks with temperature can be absorbed by the gaps between the segment without distorting the shape and direction (pattern) of the tracks

3. Here we have;

\alpha _L = Coefficient of linear expansion of iron = 12 × 10⁻⁶ /°C

ΔT = Temperature change = 27°C - 3°C = 24°C

L = Height of the Eiffel Tower = 352 meters

∴ ΔL = L × \alpha _L ×ΔT = 352 × 12 × 10⁻⁶ × 24 = 0.101376 m

Therefore, the height of the Eiffel Tower changes from 352 m to about 352.101376 m each year, with an average change in height experienced each year = 0.101376 m

4. Here, we have

L = 13.0 ft

ΔL = 1 in.

\alpha _L = 30 × 10⁻⁶ /°C

ΔT = Required temperature change

From  \frac{\Delta L}{L} = \alpha _L \Delta T

\Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{1}{156 \times 30 \times 10^{-6}} = 213.675^{\circ}C

5. Here, we have;

L = \frac{\Delta L}{\alpha _L \Delta T}

∴ L = 1/(150×25 × 10⁻⁶) = 266.67 m

The bars original length = 266.67 m

6. Here we have;

\alpha _L = \frac{\Delta L}{L \times \Delta T}

Where:

ΔL = 3.00 - 3.002 = 0.002 m

L = 3.00 m

ΔT = 110°C - 30°C = 80°C

∴ \alpha _L = 0.002/(3.00 × 80) = 8.33 × 10⁻⁶ /°C

7. Here we have;

ΔL = L × \alpha _L ×ΔT = 3 × 8.33 × 10⁻⁶ × 210 = 0.00525 m

Therefore, final length = 3.00 m + 0.00525 m = 3.00525 m

Since 3.00525 m < 3.017 m hence the alloy meets the requirement.

8. Here, we have

L = 3.2 m

ΔL = 0.5 m

ΔT = 84°C - 24°C = 60°C

∴ \alpha _L = 0.5/(3.2 × 60) = 1.95 × 10⁻³ /°C

The coefficient of linear expansion of the material from which the rod is made = 1.95 × 10⁻³ /°C

9. Here, we have

Length of steel girder, L = 32.10 m

ΔT = 8°C - 22°C = -14°C

\alpha _L = 12 × 10⁻⁶ /°C

ΔL = L × \alpha _L ×ΔT

Hence ΔL = 32.1 × 12 × 10⁻⁶× -14 = -0.0054 m

New length = 32.1 - 0.0054 = 32.095 m

10. Here we have;

ΔL = 92.6 cm - 123 cm  = -30.4 cm

\alpha _L = 2.0 × 10⁻⁵ /°C

L = 123 cm

∴ \Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{-30.4}{123 \times 2.0 \times 10^{-5}} = -12357.724^{\circ}C

Therefore, the temperature will be 200 - 12357.724 = -12157.72°C.

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