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photoshop1234 [79]
4 years ago
5

Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,0

00 km from you, and you want to know your position to 2m, what percentage uncertainity is required?
Physics
1 answer:
o-na [289]4 years ago
8 0

Answer:

0.00001 %

Explanation:

The distance from satellite = 20000 km

Position range = 2 m

The percentage uncertainty is given by dividing the distance from satellite by the position range of desired accuracy.

Percentage uncertainty is given by

\dfrac{2}{20000\times 10^3}\times 100=0.00001\ \%

The percentage uncertainty that is required is 0.00001 %

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The one on the left

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Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

R = Rydberg constant, R=1.097\times 10^7\ m^{-1}

\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

Solving above equation we get the value of final n is,

n_f=2.04

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n_f=2

So, it will relax in the n = 2. Hence, this is the required solution.        

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A wire 54.6 cm long carries a 0.480 A current in the positive direction of an x axis through a magnetic field with an x componen
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Top answer · 1 vote

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Blood velocity is not the same in all areas. In the capillaries it is where there is less speed, while in arteries and veins it is quite similar. This is due to the cross-sectional area of ​​each of the vessels. It is a mistake to think of a vein, artery or capillary individually. We have to put them all together to see that the total area of ​​the capillaries is 100 times larger than that of the arteries or veins. Blood flowing through arteries or veins is going faster because there is less area.

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