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3 years ago

A dog runs 3 miles and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Physics

1 answer:

Inga [223]3 years ago

6 0

Complete question:

A dog runs 3 miles east and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Answer:

The total distance covered by the dog is 7 miles

The displacement of the dog is 1 mile west

Explanation:

Given;

initial position of the dog = 3 miles east

final position of the dog = 4 miles west

time of motion, t = 6 hours

The total distance covered by the dog is given as;

Total distance = 3 miles + 4 miles = 7 miles

The displacement of the dog is given as;

displacement = final position of the dog - initial position of the dog

displacement = 4 miles west - 3 miles east = 1 mile west

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umka2103 [35]

Answer:

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and

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Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

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