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2 years ago

How many nanoseconds are there in 1.90 yr ? Express your answer using three significant figures.

1 answer:

5 0

   (1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)

   = (1.9 x 365.24 x 86,400 x 10⁹) nanosec

   = 6.00 x 10¹⁶ nanoseconds

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A solenoid of length 5.0 cm has a cross sectional area of 1.0 cm2 and has 300 uniformly spaced turns. The solenoid is wrapped in

OleMash [197]

The calculated mutual inductance is 8.544 x 10⁻⁵ H.

Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.

Length of the solenoid= 5.0 cm

Area of cross-section=1.0 cm²

no of spaced turns=300 turns

turns of insulated wire=180 turns

Mutual inductance (M) = μ₀μr N1N2 A/ L

=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5

=79.12 x 10⁻¹⁰ x 54000 / 5

=8.544 x 10⁻⁵ H

hence, the mutual inductance is 8.544 x 10⁻⁵ H.

Learn more about Mutual inductance here-

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4 0

9 months ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

3 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to

timama [110]

Answer:

3.486 km

Explanation:

Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:

$s = \sqrt{J^2 + M^2}$