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Leto [7]
3 years ago
6

a shopper pushes a cart 40.0m south down one aisle and then turns 90.0 degrees and moves 15.0m. He then makes another 90.0 degre

e turn and moves 20.0m. Find the shoppers total displacement. (there could be more than one correct answer)
Physics
1 answer:
valentinak56 [21]3 years ago
8 0

Answer:

Explanation:

The displacement is the distnce of the shopper from the starting point.

Sum of movement along the vertical = 40-20 = 20m

Movement along the horizontal (x direction) = 15.0m

Displacement will be gotten using the pythagoras theorem.

d = √20²+ 15²

d = √400+225

d = √625

d = 25.0m

Hence the shoppers total displacement is 25.0m

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F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

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Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

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F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

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