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e-lub [12.9K]
3 years ago
11

An object moves from point A to B to C to D and finally to A

Physics
1 answer:
Nataly [62]3 years ago
5 0

here's the solution,

The <em>radius</em> of the circle =<u> 3 km</u>

distance covered = <em>circumference</em> of the circle,

So, Circumference :

=》

2\pi r

=》

2 \times 3.14 \times 3

=》

18.84

(a). Distance covered by moving object is 18.84 km

(b). 0 km

now, Displacement of the object is 0 km, because displacement is the shortest distance from stating point to the destination, but the object returns back to the starting point, hence magnitude of displacement is 0.

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Answer:

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One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-
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Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = 0.12  \times 10^{-2} \ m

x = 1.17 \ cm = 1.17 \times 10^{-2}\ m

m = 149 kg

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da = 2.28 \times 10^{-10}\ m

F_{net} = F-mg\\ \\0 = F - mg \\ \\  F = mg \\ \\ k_sx = mg \\ \\

∴

k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\  k_s = 124803.42  \ N /m

N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}

N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2

N_{chain} =  (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2

N_{chain} = 2.77 \times 10^{13}

N_{bond} = \dfrac{L}{da} \\ \\  = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}

\text{Finally; the stiffness of a single interatomic spring is:}

k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s

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\mathbf{k_{si} =45.46 \ N/m}

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pickupchik [31]

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Explanation:

The answer is false.

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where:

m is the mass of the object

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We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

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