How much farther from the charge is the 2000 v equipotential surface than the 3000 v surface?

One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t

natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = $\frac{24\ grams}{6\ x\ 10^{23}\ atoms}$ = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

$\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}$

where,

$\rho$ = density = 1.7 grams/cm³
m = mass of single atom = 4 x 10⁻²³ grams
V = volume of single atom = ?

Therefore,

$V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}$

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

$V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}$

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

Learn more about density here:

brainly.com/question/952755

7 0

2 years ago

Calculate the moment of inertia for each scenario: (a) An 80.0 kg skater is approximated as a cylinder with a 0.140 m radius. (b

Zina [86]

Answer:

a) the moment of inertia is 0.784 Kg*m²

b) the moment of inertia is with arms extended is 1.187 Kg*m²

c) the angular velocity in scenario (b) is 4.45 rad/s

Explanation:

The moment of inertia is calculated as

I = ∫ r² dm

since

I = Ix + Iy

and since the cylinder rotates around the y-axis then Iy=0 and

I = Ix = ∫ x² dm

if we assume the cylinder has constant density then

m = ρ * V = ρ * π R²*L = ρ * π x²*L

therefore

dm = 2ρπL* x dx

and

I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 = mR² /2

therefore

I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²

b) since the arms can be seen as a thin rod

m = ρ * V = ρ * π R²*L = ρ * π R²*x

dm =ρ * π R² dx

I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)

= ρ * π R²*2*L³/24 = mL²/12

therefore

I skater 2 = I1 + I skater = mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²

c) from angular momentum conservation

I s2 * ω s2 = I s1 * ω s1

thus

ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s

4 0

3 years ago

A well insulated and perfectly sealed room is installed with 4 HP fan to circulate air for 1.5 hours. Determine the increase in

marissa [1.9K]

"Increase in internal energy = energy consumed by fan " as the room is perfectly insulated and sealed.

energy consumed by fan = power x time
power of fan = 4HP = 4 x 745.7 W = 2982.8 W
time = 1.5 hours = 1.5 x 60 x 60 = 5400 s

energy consumed by fan = 2982.8 x 5400 = 16107120 J = 1.61x10^7 J

Increase in internal energy is 1.61x10^7 J

7 0

3 years ago

During which section was the person moving at the highest speed? What was their speed? Defend your answer.

alexandr1967 [171]

Where's the person that the question is talking about?

5 0

3 years ago

Read 2 more answers

a skier speeds along a flat patch of snow, and then flies horizontally off the edge at 16.0 m/s. He eventually lands on a straig

geniusboy [140]

Answer:

1.63 s

Explanation:

The skier lands on the sloped section when the direction of its velocity is exactly identical to that of the slope, so at $45.0^{\circ}$ below the horizontal.

This occurs when the magnitude of the vertical velocity is equal to the horizontal velocity (in fact, $\tan \theta =\frac{|v_y|}{v_x}$ , and since $\theta=45.0^{\circ}, tan \theta = 1$ and so $|v_y| = v_x$ .