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Diano4ka-milaya [45]

3 years ago

9

A group of runners complete a 26.2 mile marathon in 3.4 hours. The distance between the start and finish lines is 12.2 miles. Wh

at was the average velocity for the run?

2 answers:

NNADVOKAT [17]3 years ago

8 0

26.2/3.4 would be the average velocity for the run.

7.7 miles/hr

wlad13 [49]3 years ago

5 0

Answer:

$\vec v = 3.6 mph$

Explanation:

As we know that the total path length of the Marathon is given as

$L = 26.2 miles$

while the straight distance between start and finish lines is given as

$d = 12.2 miles$

so here we can say that the displacement is the straight line distance between initial and final position which is 12.2 miles

so it is given as

$\vec d = 12.2 miles$

now the average velocity is defined as the ratio of displacement and time

$\vec v = \frac{\vec d}{time}$

$\vec v = \frac{12.2 miles}{3.4 hr}$

$\vec v = 3.6 mph$

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<u>r < a:</u>

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<u>r = a:</u>

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<u>a < r < b:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>r = b:</u>

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

<u>b < r < c:</u>

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<u>r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

<u>r < c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u /> $E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$ <u />

<u>At r = b:</u>

<u /> $E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$ <u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

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<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>At r = c:</u>

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<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

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so, the frictional force will be:

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