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4 years ago

What is the relationship between the atoms of elements,its isotopes and its ions

1 answer:

6 0

An element refers to a collection of atoms having the same number of protons and electrons (an atomic number). In each element there is a different atomic number due to a different amount of protons in the nucleus.

An isotope is a variation of an element that contains a different number of neutrons, therefore adding weight to the atom.

An ion is a charged atom, and its charge shows how many electrons it needs to gain or lose in order to become stable.

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In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 4.32 x 104-m length of wire

elena55 [62]

Answer:

Explanation:

Length, l = 4.32 x 10^4 m

speed, v = 7.07 x 10^3 m/s

magnetic field, B = 5.81 x 10^-5 T

The formula for the motion emf is given by

e = B x v x l

e = 5.81 x 10^-5 x 7.07 x 10^3 x 4.32 x 10^4

e = 17745.1 V

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3 years ago

A 15-kg ball is tossed up into the air. The ball is 2 meters off the ground traveling 4 m/s. What is the potential energy? A. 29

Sladkaya [172]

Answer: 0j

Explanation:

At that point potential energy is zero and kinetic energy is maximum.. P. E=mgh=0

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4 years ago

What The BEST GAME?????/ a: Roblox b; Mincraft c; Fortnite

Maksim231197 [3]

Answer:

A,C

Explanation:

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3 years ago

Read 2 more answers

What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An

GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}$

$E=\frac{2m_{0}c^}2}{\sqrt{3}}$

$E=\frac{2\times0.001\times9\times10^{16}}{1.732}$

$E=1.04\times10^{14}J$

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, $E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV$

4 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$