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2 years ago

9

NEED HELP!!!!! 10 POINTS

2 answers:

castortr0y [4]2 years ago

6 0

Answer: D.

Explanation:

satela [25.4K]2 years ago

4 0

The answer is D I’m pretty sure

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If light needs matter , how doe it move through space? this is my first question.

Assoli18 [71]

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3 years ago

A 7382 N piano is to be pushed up a(n) 2.16 m frictionless plank that makes an angle of 22.7 ◦ with the horizontal. Calculate th

lubasha [3.4K]

Answer:

Work done, W = 6153.31 Joules

Explanation:

It is given that,

Weight of piano, W = F = 7382 N

It is pushed 2.16 meters friction less plank

The angle with horizontal, $\theta=22.7^{\circ}$

When the piano slide up plank at a slow constant rate. The y component of force is taken into consideration. The net force acting on it is given by :

$F_y=F\ sin\theta$

Work done is given by :

$W=F_y\times d$

$W=F\ sin\theta \times d$

$W=7382 sin(22.7) \times 2.16$

W = 6153.31 Joules

So, the work done in sliding the piano up the plank is 6153.31 Joules. Hence, this is the required solution.

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2 years ago

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-

choli [55]

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

$\delta = 7.87 \ g/cm^3$

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

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

∴

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

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

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

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

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

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

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

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

$k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)$

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

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2 years ago

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2 years ago

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3 years ago