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3 years ago

9

A runner is training and runs past meter mark 65 and starts a stopwatch. After the runner passes meter mark 15 the stopwatch sto

ps at 8 seconds.
What was the runners velocity?
Question options:
-6.25 m/s
6.25 m/s
8.13 m/s
-1.88 m/s

1 answer:

denis-greek [22]3 years ago

7 0

Answer:

6.25 m/s

Explanation:

Given parameters

Initial mark = 65

Final mark = 15

time taken = 8s

Unknown:

Velocity of the runner = ?

Solution:

Velocity is the displacement divided by time;

Velocity = $\frac{displacement}{time}$

Displacement = Initial mark - final mark = 65 - 15 = 50m

Now, insert the parameters and solve;

Velocity = $\frac{50}{8}$ = 6.25m/s

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Explanation:

Given

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

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Answer:

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and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

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on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

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Answer:

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