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4 years ago

He goes 12km north 6km east 12km south 3km west what is the distance and displacement

Physics

1 answer:

Hunter-Best [27]4 years ago

6 0

Distance is 33 km, displacement is 3 km east

Explanation:

- Distance is a scalar quantity equal to the total length of the path covered by an object during its motion, regardless of its direction.

- Displacement is a vector connecting the starting position of motion of an object to its final position. It is a vector quantity, therefore it has both magnitude and direction.

In this problem, the motion of the person is:

12 km north

6 km east

12 km south

3 km west

Therefore, the distance is:

$d=12+6+12+3=33 km$

The displacement instead can be calculated by comparing the final position with the initial position. We notice that:

- the person goes 12 km north and 12 km south, so he didn't move in the north-direction

- The person goes 6 km east and 3 km west, so his final position is 3 km east

Therefore, the displacement is 3 km east.

Learn more about distance and displacement:

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4 years ago

Read 2 more answers

You are setting up a physics experiment to measure the electric field of a dipole along its axis. At what distance would you exp

snow_lady [41]

Answer:

y = 1.73 √s

Explanation:

For this question, let's look for the complete formula for the elective field of a dipole and then compare with the approximate formula.

A dipole is two charges of equal magnitude and different sign separated a distance 2, the field on axes at the midpoint is

E₁ = E₂ = k q² / r²

For distance we use Pythagoras' theorem

r² = y² + s²

The total electric field is

E = 2 E₁ cos θ

The field perpendicular to the dipole axis is canceled, let's use trigonometry

cos θ = s / r

Let's replace

E = 2 k q² / (y² + s²) a / √(y² + s²)

E = 2 q s / (y² + s²)^{3/2}

This is the exact formula.

The approximate formula is

E’= 2 q s / y³

If we relate these two formulas

E’/ E = (y² + s²).^{3/2}/y³

We see that the error in the distance propagates in an error for the electric field, they ask us that the uncertainty be 5% (er = 0.05)

The approximate formula is the measured value and the exact formula is the actual calculated value, so the relative uncertainty is

E’= E (y² + s²).^{3/2} / y³

ΔE ’= dE’ /dy Δy + dE’/ds Δs + dE’ /dE ΔE

The last term is considered zero since the value is exact

dE’/ dy = (y² + s²).^{3/2} (-3y⁻⁴) + y⁻³ 3/2 (y² + s²).^{1/2} 2y

dE ’/ dy = -3 (y² + s²).^{3/2}/y⁴ - 3y (y2 + s2).^{1/2} /y³

dE ’/ ds = 3/2 (y² + s²).^{1/2} 2s/y³

dE'/ds = 3s (y²+s²).^{1/2} /y³

ΔE’= E [+3 (y² + s²).^{3/2}/y⁴ + 3y (y2 + s2).^{1/2} /y³] Δy

+ [3s (y²+s²).^{1/2} /y³] Δs

ΔE’/E’ = Δy [3y - 3 / (y² + s²)] + Δs [3s / (y² + s²)]

ΔE’/E' = 3Δy [(1- / (y² + s²)] + 3Δs s / (y² + s²)

In general the distance and is measured with a tape measure, large value with an uncertainty of Δy = 0.1 cm and the distance between the charge is measured with a caliper Δs = 0.05 cm

Let's replace the values

0.05 = 0.1 3[1 – 1/ (y² + s²)] + 0.05 3s /(y² + s²)

This is the formula of the error between the approximate field and the exact field, so that the error is at 0.05, the first term must be eliminated by which y >> s

0.05 = 0.05 3s / y²

1 = 3s / y ²

y = √3s

y = 1.73 √s

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