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Alex Ar [27]
3 years ago
11

Zamir and Talia raced through a maze. Zamir walked 2 m north, 2 m east, 4 m south, 2 m east, 4 m north, 2 m east, 3 m south, 4 m

east and 4 m north. Talia walked 2 m north, 6 m east, 3 m south, 4 m east, and 4 m north. Compare their displacements:
Physics
2 answers:
Alja [10]3 years ago
6 0

Answer : Zamir's displacement and Talia's displacement is equal.

Explanation :

Displacement is explained to be the changing position of an object.

Zamir covers total distance 27 m and Talia covers total distance 19 m but  Zamir's initial and final position and Talia's initial and final position is same.

So, we can say that Zamir's displacement and Talia's displacement is equal.

Allisa [31]3 years ago
6 0

Zamir’s displacement is equal to Talia’s displacement.

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lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air y at time t is given by

y=100\,\mathrm m-\dfrac g2t^2

Let d be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}

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Why is it impossible to build a machine that produces more energy than it uses?
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A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
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Answer:

-4*10⁴ units.

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As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must  still remain to be zero.

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
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Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

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T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

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Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

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