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Montano1993 [528]

1 year ago

13

A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

what is the magnitude of the joggers net displacement?

1 answer:

Annette [7]1 year ago

6 0

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

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1 year ago

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klemol [59]

Answer:

(a) 2.8 V

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Now, impedance is given as:

$Z=\sqrt{X_L^2+X_C^2}\\\\Z=\sqrt{(49)^2+(28)^2}\\\\Z=\sqrt{3185}=56.4\ \Omega$

Current across the circuit is given as:

$I=\frac{V}{Z}\\\\I=\frac{12}{56.4}=0.2\ A$

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$V_R=IR\\\\V_R=0.2\times 14=2.8\ V$

Therefore, the voltage across resistor is 2.8 V.

(b)

Voltage across the capacitor is given as:

$V_C=IX_C\\\\V_C=0.2\times 28=5.6\ V$

Therefore, the voltage across the capacitor is 5.6 V.

(c)

Voltage across the inductor is given as:

$V_L=IX_L\\\\V_L=0.2\times 49=9.8\ V$

Therefore, the voltage across the inductor is 9.8 V.

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Answer:

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