Answer:
<h2>
ΔE = -1,382,800Joules</h2>
Explanation:
The internal energy of the system according to the thermodynamics law is the difference between the heat released by the system and work done to the system. Mathematically;
ΔE = q - W where;
ΔE is the change in internal energy of the syatem
q is the heat released by the system
W is the work done to the system
Given W = 4.50×10² kcal and q = 5.00×10² kJ
Since 1cal = 4.184Joules
4.50×10² kcal = (4.50×10²×10³)×4.184
W = 4.50×10² kcal = 1,882,800 Joules
q = 5.00×10² kJ = 5.00×10⁵Joules
q = 500,000Joules
Since ΔE = q - W
ΔE = 500,000 J - 1,882,800 J
ΔE = -1,382,800Joules
Hence, the change in the internal energy of the system is -1,382,800Joules
Answer: Review your data. ...
Calculate an average for the different trials of your experiment, if appropriate.
Make sure to clearly label all tables and graphs. ...
Place your independent variable on the x-axis of your graph and the dependent variable on the y-axis.
Hope this helps!!
Answer:
force=337.5N
accelertion=5.63ms^-2
Explanation:
please find the answers in the pictures above maam or sir.
Answer:
ΔR = 9 s
Explanation:
To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value
The given expression is R = 2A / B
the uncertainty is ΔR = | | ΔA + | | ΔB
we look for the derivatives
= 9 / B
= 9A ( )
we substitute
ΔR = ΔA + ΔB
the values are
ΔA = 2 s
ΔB = 3 s
ΔR = 2 + 3
ΔR = 1.636 + 7.14
ΔR = 8,776 s
the absolute error must be given with a significant figure
ΔR = 9 s