Answer:
equilibrium position.
Explanation:
In simple harmonic motion , velocity v(t) is given by,
v(t) = -ω A sin(ωt + φ)
where
ω = angular velocity of the corresponding circular motion
A = amplitude
t = time
φ = the initial angle of the corresponding circular motion when the motion begin.
v (t) get maximized when sin value is maximized , i.e. sin 
=1
The particle has maximum speed when it passes through the equilibrium position.
Answer:
To summarize and analyze data with both a crosstabulation and charting, Excel typically pair PivotCharts with PivotTables (option b)
Explanation:
Pivot Tables allows us to deal with a high volume of data. This kind of tables are mainly designed to summarize, analyze and explore data in a simple way. Meanwhile, Pivot Charts gives the possibility to visualize the summarizing data provided in PivotTables, enabling the detection of trends, correlations,comparisons and so on. Because of that Pivot tables and pivot charts are usefull and complementary tools of Excel.
Summarizing, the proper option is b.
I know that its not the second law. I'm almost positive its the first one. Please let me know if I'm wrong. This sentence makes no sense when you put it with the third law. So, the first law is my guess...
Explanation:
Formula for steady flow energy equation for the flow of fluid is as follows.
![m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w](https://tex.z-dn.net/?f=m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%5D%20%2B%20z_%7B1%7Dg%5D%20%2B%20q%20%3D%20m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%20%2B%20z_%7B1%7Dg%5D%20%2B%20w)
Now, we will substitute 0 for both 
and 
, 0 for w, 334.9 kJ/kg for 
, 2726.5 kJ/kg for 
, 5 m/s for 
and 220 m/s for 
.
Putting the given values into the above formula as follows.
![1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0](https://tex.z-dn.net/?f=1%20%5Ctimes%20%5B334.9%20%5Ctimes%2010%5E%7B3%7D%20J%2Fkg%20%2B%20%5Cfrac%7B%285%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%20q%20%3D%201%20%5Ctimes%20%5B2726.5%20%5Ctimes%2010%5E%7B3%7D%20%2B%20%5Cfrac%7B%28220%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%200)
q = 6597.711 kJ
Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.
There is no difference. 5 is the same as 5.0
