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3 years ago

6

Which part of the steam engine does the work?

2 answers:

erik [133]3 years ago

6 0

Imagine living off nothing but coal and water and still having enough energy to run at over 100 mph! That's exactly what a steam locomotive can do. Although these giant mechanical dinosaurs are now extinct from most of the world's railroads, steam technology lives on in people's hearts and locomotives like this still run as tourist attractions on many heritage railways.

Steam locomotives were powered by steam engines, and deserve to be remembered because they swept the world through the Industrial Revolution of the 18th and 19th centuries. Steam engines rank with cars, airplanes, telephones, radio, and television among the greatest inventions of all time. They are marvels of machinery and excellent examples of engineering, but under all that smoke and steam, how exactly do they work?

vazorg [7]3 years ago

5 0

The piston is correct

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2 years ago

Question 4 - If Angelica starts out at 30m/s, and in 16 s speeds up to 84 m/s, what is her acceleration?

Triss [41]

Answer:

C. $3.375 m/s^2$

Explanation:

The acceleration of an object can be found using the equation:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time it takes for the velocity to change from u to v

In this problem:

u = 30 m/s is the initial velocity of Angelica

v = 84 m/s is the final velocity

t is the time

Substituting into the equation, we find the acceleration:

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3 years ago

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2 years ago

Early in the morning, when the temperature is 4.5 ∘C, gasoline is pumped into a car's 53-L steel gas tank until it is filled to

Lynna [10]

Answer:

$\Delta V=1.0911\ L$

Explanation:

Given:

Initial temperature, $T_i=4.5^{\circ}C$

initial volume, $V_i=53\ L$

final temperature, $T_f=27^{\circ}C$

volumetric coefficient of thermal expansion for gasoline, $\beta_g=950\times 10^{-6}\ K^{-1}$

volumetric coefficient of thermal expansion for steel, $\beta_s=35\times 10^{-6}\ K^{-1}$

<u>Now the increment in the volume of the container:</u>

$\Delta V_s=V_i.\beta_s.\Delta T$

$\Delta V_s=53\times 35\times 10^{-6}\times (27-4.5)$

$\Delta V_s=41737.5\times 10^{-6}\ L$

<u>and the increment in the volume of the gasoline:</u>

$\Delta V_g=V_i.\beta_g.\Delta T$

$\Delta V_g=53\times 950\times 10^{-6}\times (27-4.5)$

$\Delta V_g=1132875\times 10^{-6}\ L$

<u>Hence the difference in the increment of the two volumes:</u>

$\Delta V=\Delta V_g-\Delta V_s$

$\Delta V=(1132875-41737.5)\times 10^{-6}$

$\Delta V=1.0911\ L$ of the gasoline spills out of the container in this case.

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3 years ago

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Ostrovityanka [42]

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2 years ago