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Katarina [22]
3 years ago
8

A train moves at constant velocity of 90km/h. How far will it go in 0.25h.

Physics
1 answer:
enot [183]3 years ago
8 0
In a quarter of an hour it will travel 1/4 x 90 = 22.5Km
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Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in
Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

L = \frac{N^2 \mu_0 A}{l}

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

3 0
2 years ago
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:

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