When a puddle dries up what are the particles really doing

A man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed

klemol [59]

He waits until night and goes through the door with the magnifying glass. The sun is not out so it can not instantly fry him.

4 0

3 years ago

A piece of curved glass has a radius of curvature of r = 10.8 m and is used to form Newton's rings, as in the drawing. Not count

MAVERICK [17]

Answer:

Radius of the outer most dark fringe is 2.65 cm

Solution:

As per the question:

Radius of curvature of the glass, r = 10.8 m

No. of dark fringes, n = 100

Wavelength of light, $\lambda = 652\ nm = 652\times 10^{- 9}\ m$

Now,

To calculate the radius R of the outermost ring:

Radius of the dark fringe of nth order is given by:

$R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}$

$R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm$

7 0

4 years ago

Read 2 more answers

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

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

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

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

∴

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

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

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

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

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

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

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

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

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

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

4 0

3 years ago

An object slides down a frictionless inclined plane. At the bottom, it has a speed of 9.80 m/s. What is the vertical height of t

Papessa [141]

The vertical height of the given plane is 4.9 m.

The given parameters:

<em>speed of the object at the bottom of the ramp, v = 9.8 m/s</em>

The vertical height of the plane is calculated by applying principle of conservation of mechanical energy as follows;

$P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{9.8 \times 9.8}{2 \times 9.8} \\\\h = 4.9 \ m$

Thus, the vertical height of the given plane is 4.9 m.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0

3 years ago

What is the relationship between electric field lines and equipotential lines that you observed in doing the lab

lidiya [134]

Answer:

Explained below

Explanation:

Generally speaking, we know in physics that Electric field lines are lines which usually start at positive charges and deflect away from them to terminate at the negative charges. Meanwhile Equipotential lines are lines that are used to connect points located on the same electric potential.

Finally, in conclusion, electric field lines are usually lines that go through in a perpendicular manner across every equipotential lines.

8 0

3 years ago