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3 years ago

Why must indirect evidence be used to study the structure of atoms?

1 answer:

8 0

Indirect evidence must be used to study the structure of atoms, because the direct evidence affects the position and energy of the particles of the atomic structure.

For example, imagine you use a light ray to find the location of an electron in an atomic structure, the current location of the electron you are now seeing is as a result of the light ray, and the electron has changed it's initial position to a new position as a result of the effect of the light ray.

So indirect means would probably give more accurate answers about the atomic structure than that of the direct means.

An example of the indirect means would be for example setting up a mathematical equation or model of the atomic structure and trying to provide solution for the model or equation.

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Tell me the max amount you should owe on this card.

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Answer:

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Explanation:

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3 years ago

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3 years ago

Read 2 more answers

During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.

Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

$Q = \frac{\Delta P \pi r^4}{8\eta l}$

Where:

$\eta_i =$ are the viscosities of the concrete before and after the increase

l = Length of the vessel

$r_1, R_2$ = Radio of the vessel before and after the increase

$\Delta P$ = Change in the pressure

$Q_{1,2} =$ The rates of flow before and after he increase

Our values are given as:

$Q_2 = 10Q_1 \rightarrow$ 10 times her resting rate

$\eta_2 = 0.95\eta_1$ 95% of its normal value

$\Delta P_2 = 1.5\Delta P_1$ Increase of 50%

Plugging known information to get

$Q_1 = \frac{\Delta P \pi r^4}{8\eta l}$

$Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4$

$r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}$

$r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}$

$r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}$

$r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}$

$r_2 = 1.586r_1$

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

7 0

4 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$