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Tamiku [17]
2 years ago
7

What is the correct displacement for the following vectors: 4 km south, 2 km north, 5 km south, and 5 km north? 2 km south 16 km

total 11 km west 6 km south
Physics
1 answer:
lesantik [10]2 years ago
6 0

<u>Answer:</u>

The correct answer option is: 2 km South.

<u>Explanation:</u>

We are to find out the displacement for the given vectors:

- 4 km south,

- 2 km north,

- 5 km south; and

- 5 km north

So starting off with the vector going 4 km south we have a displacement of 4 km south. Then it goes 2 km so we are left with 4 - 2 = 2 km south.

Next, it goes 5 km south so we will have a displacement of 2 + 5 = 7 km south. Then again it goes 5 km north so we are left with a displacement of 7 - 5 = 2 km South.

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Answer:

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Explanation:

(a)

Distance between the maximum and the minimum of the  wave = 2A ............ Equation 1

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Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

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(b)

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Given: f = 4.31 Hz

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T = 0.23 s.

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Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

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(c)

If  1.21 m contains  1 cycle,

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Answer:

\boxed {\boxed {\sf D. \ 6.02*10^{23} \ atoms}}

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Regardless of the particles, there will always be <u>6.02*10²³</u> (also known as Avogadro's Number) particles in one mole of a substance.

Therefore, the best answer for 1 mole of titanium is D. 6.02*10²³ atoms.

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