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a_sh-v [17]
2 years ago
12

A 58.7-g sample of nickel (s = 0.443 J/(g ∙ °C)), initially at 147.4°C, is placed in an insulated vessel containing 184.0 g of w

ater (s = 4.18 J/(g ∙ °C)), initially at 29.9°C. Once equilibrium is reached, what is the final temperature of the metal–water mixture?
Physics
1 answer:
kari74 [83]2 years ago
3 0

Answer:

Explanation:

Let t be the final temp

temp. drop of nickel=147.4-t

temp. gain by water=t-29.9

in equilibrium

Heat lost=Heat gained

58.7×0.443×(147.4-t)=184×4.18×(t-29.9)

58.7×0.443×147.4-58.7×0.443×t=184×4.18×t-184×4.18×29.9

     3833.00434 -26.0041 t=769.12t-22996.688

(769.12 +26.0041)t= ?

find t

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There are three questions here:


A. The possible daily output of this process if there is 8 processing time each day?


the time it takes to assemble a chair in seconds = A + B + C + J + K + L + X + Y + Z  

which is equal to = 38 + 34 + 35 + 32 + 30 + 34 + 22 + 18 + 20 =263 per chair



Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60


= 480 x 60 = 28,800 seconds is available in an 8 hr day.



28,800 / 263 =109.5057034220532


Therefore, It is possible to make 109 chairs in an 8 hour day.



B. Given your output rate in above, what is the efficiency of the process?


The time it takes to assemble a chair in seconds = A+ B + C + J + K + L + X + Y + Z.


= 34 + 34 + 34 + 30 + 30 + 30 + 22 + 18 + 20 = 252.6315789473684 per chair


A total of 252 per chair


Processing hours = 8 hours x 60 minutes x 60 seconds


= 8 x 60 x 60


= 480 x 60 = 28,800 seconds is available in an 8 hour day


= 28,000 / 252 = 114.2857142857143114 chairs


= 109/114 x 100 = 95.6140350877193


Therefore, the efficiency of making the chair is 95.61%


C. What is the flow time of the process?


Flow time is the period that it takes a completed chair to flow through the process from the beginning assemblage step to the last step. Take note that ABC and JKL are parallel legs in the process, and as a result both do not include to flow time to the procedure. In addition, Flow time comprises both pass time and run time at each position in the procedure.

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Explanation:

Below is an attachment containing the solution.

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An actor who memorizes his lines word-for-word has demonstrated _____ learning.
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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

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(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

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