Answer:
The answer is "[tex]-121\ \frac{KJ}{Kg}[/tex]".
Explanation:
Please find the correct question in the attachment file.
using formula:
[tex]\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \ or \ \ \frac{RT_1 -RT_2}{n-1}\\\\[/tex]
[tex]W =\frac{R(T_1 -T_2)}{n-1}\\\\[/tex]
[tex]=\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\[/tex]
Americans spent approximately __ cents of each dollar on the purchase of energy?
Answer:
14 cents
Explanation:
1 trillion US dollar
What are motion study principles? How are they classified?
Frederick Taylor created a series of guidelines known as the "motion study principles" to boost productivity at work. Time study, motion study, fatigue study, etc technique conduct research are the four divisions into which they fall.
What is motion study principles?Motion studies are the study of actions like lifting, putting things down, standing, changing direction, and other actions that are made when performing routine work.
In order to execute the task effectively in less time, unnecessary moves are attempted to be removed. The following three headings comprise the fundamentals: Making use of the human body. The layout of the workplace. Machinery and tools design.
Motion study principles will be the which will help in designing or putting films into a particular order. Also how the routine works are being made is suggested.
Learn more about motion study, here:
https://brainly.com/question/30164973
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how to update android 4.4.2 to 5.1 if there isnt any update available
Answer:
You cant update it if there isin´t any update available lol
Explanation:
When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note that the rope runs around the pulley B and a pin attached to block C, as indicated.
Answer:
The answer is "2 m/s".
Explanation:
The triangle from of the right angle:
[tex]\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant[/tex]
Differentiating the above equation:
[tex]\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\[/tex]
[tex]\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\[/tex]
[tex]= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}[/tex]