Because Earth orbits around the sun, causing the position to seemingly change as the Earth moves. Orion can't be seen in summer because it appears when the sun is out. The reason it doesn't appear at night is because the earth is titled differently on it's axis then in, say winter.
I hope this helps you out.
Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=![=\frac{450}{1000}=0.45 kg](https://tex.z-dn.net/?f=%3D%5Cfrac%7B450%7D%7B1000%7D%3D0.45%20kg)
Where 1 kg=1000 g
Frequency of oscillation=![\nu=1.2Hz](https://tex.z-dn.net/?f=%5Cnu%3D1.2Hz)
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=![E=\frac{1}{2}m\omega^2A^2](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7Dm%5Comega%5E2A%5E2)
Where
=Angular frequency
A=Amplitude
![\pi=\frac{22}{7}](https://tex.z-dn.net/?f=%5Cpi%3D%5Cfrac%7B22%7D%7B7%7D)
Using the formula
![0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2](https://tex.z-dn.net/?f=0.51%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.45%282%5Ctimes%20%5Cfrac%7B22%7D%7B7%7D%5Ctimes%201.2%29%5E2A%5E2)
![A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398](https://tex.z-dn.net/?f=A%5E2%3D%5Cfrac%7B2%5Ctimes%200.51%7D%7B0.45%5Ctimes%20%282%5Ctimes%20%5Cfrac%7B22%7D%7B7%7D%5Ctimes%201.2%29%5E2%7D%3D0.0398)
![A=\sqrt{0.0398}=0.199](https://tex.z-dn.net/?f=A%3D%5Csqrt%7B0.0398%7D%3D0.199)
Hence, the amplitude of oscillation=A=0.199
Answer:
c
Explanation:
It really depends on what the stuff is. This should not be a math or physics question.
Having said that, I answer it by rounding it to 1. 1.04 is closest two one.
That is not entirely a satisfactory answer. Most people who take medications have pill splitters. They easily cut a pill in half. They can do this only if a the medicine's directions, or a pharmacist, or a doctor tells you can do this. Cutting it into quarters should never be done.
So a is really incorrect medically.
1.5 is too far away
C is probably the intended answer; it would annoy me no end if it was anything but c. If this was just a rounding question, it rounds to c. There is no medication that requires an intake of 1.04. No one could get that kind of accuracy.
2 is really too far away.
1) See attached graph
To solve this part of the problem, we have to keep in mind the relationship between current and charge:
![i = \frac{\Delta Q}{\Delta t}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7B%5CDelta%20Q%7D%7B%5CDelta%20t%7D)
where
i is the current
Q is the charge
t is the time
The equation then means that the current is the rate of change of charge over time.
Therefore, if we plot a graph of the charge vs time (as it is done here), the current at any time will be equal to the slope of the charge vs time graph.
Here we have:
- Between t = 0 and t = 2 s, the slope is
, so the current is 25 A
- Between t = 2 s and t = 6 s, the slope is
, so the current is -25 A
- Between t = 6 s and t = 8 s, the slope is
, so the current is 25 A
Plotting on a graph, we find the graph in attachment.
2) ![15 \mu C](https://tex.z-dn.net/?f=15%20%5Cmu%20C)
The relationship we have written before
![i = \frac{\Delta Q}{\Delta t}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7B%5CDelta%20Q%7D%7B%5CDelta%20t%7D)
Can be rewritten as
![\Delta Q = i \Delta t](https://tex.z-dn.net/?f=%5CDelta%20Q%20%3D%20i%20%5CDelta%20t)
This is valid for a constant current: if the current is not constant, then the total charge is simply equal to the area under the current vs time graph.
Therefore, we need to find the area under the graph.
Here we have a trapezium, where the two bases are
A = 1 ms = 0.001 s
B = 2 ms = 0.002 s
And the height is
h = 10 mA = 0.010 A
So, the area is
![Area=\frac{(0.001+0.002)\cdot 0.010}{2}=1.5\cdot 10^{-5} C = 15 \mu C](https://tex.z-dn.net/?f=Area%3D%5Cfrac%7B%280.001%2B0.002%29%5Ccdot%200.010%7D%7B2%7D%3D1.5%5Ccdot%2010%5E%7B-5%7D%20C%20%3D%2015%20%5Cmu%20C)
So, the charge is
.
Profit = 1350-1184=166 AED
Loss= 1350-166 =1184