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3 years ago

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A cross-country runner accelerates from 4 m/s to 7 m/s over the course of 5 seconds. What is the runner's average rate of accele

ration? A. 0.3 m/s/s B. 0.6 m/s/s C. 1.4 m/s/s D. 2.2 m/s/s

1 answer:

fredd [130]3 years ago

4 0

B. 0.6 for show
hopefully this works lemme know

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A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

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Answer:

$v_{max} = |d|\cdot \sqrt{\frac{k}{m} }$

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

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Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$

We need to asume here the number of free electrons in a copper conductor, at which is generally of $8.5 *10^{28}m^{-3}$

The equation to find the current is

$I = VenA$

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

$I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)$

$I= 22.11a$

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$I = \frac{Q}{t} \rightarrow Q = It$

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

$V_d = \frac{l}{t} \rightarrow l = V_d t$

The charge density is defined as

$\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}$

Replacing our values

$\lambda = \frac{22.11}{5.20*10{-4}}$

$\lambda= 4.25*10^4C/m$

Therefore the charge density in the system is $4.25*10^4C/m$

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