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ahrayia [7]
3 years ago
8

Explain why a fast moving freight train cannot be stopped quickly

Physics
1 answer:
nika2105 [10]3 years ago
8 0
A fast moving freight train can not be stopped quickly because of Newton's law of Inertia.Once a object is in motion it takes force to be stop it and it wants to keep going so it's harder to stop.
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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.

3 0
3 years ago
a crane lifts a 75 kg mass at a height of 8m . calculate the gravitational potential energy gained by the mass (g= 10N)
Bogdan [553]
Advice: Search up your questions in the Brainly search bar before asking your questions.
5 0
3 years ago
If vector A=2i+3j-k and vector B=4i+6j-2k. The angle between them will be: a) 0°b)45°c)90°d)60°​
melomori [17]

Notice that

<em>B</em> = 4<em>i</em> + 6<em>j</em> - 2<em>k</em> = 2 (2<em>i</em> + 3<em>j</em> - <em>k</em>) = 2<em>A</em>

so both vectors point in the same direction and the angle between them is (A) 0°.

7 0
3 years ago
2) A 20 kg mass moving at a speed of 3 m/s is stopped by a constant force of 15 N. How many seconds must the force act on the ma
hammer [34]

Take the object's starting direction of motion to be the positive direction, so that a stopping force acts in the opposite direction. By Newton's second law, the object undergoes an acceleration <em>a</em> such that

-15 N = (20 kg) <em>a</em>

Solve for <em>a</em> :

<em>a</em> = - (15 N) / (20 kg) = -0.75 m/s²

The object's velocity <em>v</em> at time <em>t</em> is then given by

<em>v</em> = 3 m/s + (-0.75 m/s²) <em>t</em>

so the time it takes for the object to slow to a rest is

0 = 3 m/s + (-0.75 m/s²) <em>t</em>

<em>t</em> = (3 m/s) / (0.75 m/s²) = 4.0 s

3 0
3 years ago
Before collecting a specimen of urine or feces, the nursing assistant asks the nurse or consults the lab for?
bulgar [2K]

Before collecting a specimen of urine or feces, the nursing assistant asks the nurse or consults the lab for which storage and delivery method to use.

There are different tests that need to be performed on a urine or feces specimen. It depends on the patients that which test would be asked for him to be done.

The storage and delivery methods for urine and feces in the case of different tests has to be consulted by the nurse or consultant.

Incase of an emergency test, whose results are immediately required the method of delivery is a fast mode one. For patients that have a mild disease or are not at risk, other reliable methods can be used. Each specimen should be properly labeled for proper checking.

To learn more about urine specimen, click here:

brainly.com/question/28329488

#SPJ4

8 0
2 years ago
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