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nikitadnepr [17]
3 years ago
14

Atoms seldom exist as independent particles in nature because...

Physics
1 answer:
ExtremeBDS [4]3 years ago
3 0
<h3><u>Answer;</u></h3>

C) atoms are more stable when they combine with other atoms

<h3><u>Explanation</u>;</h3>
  • Atoms are the smallest particles of an element that can take part in a chemical reaction.
  • <em><u>Atoms seldom exist as independent particles in nature. Nearly all substances are made up of combinations of atoms that are held together by chemical bonds, which are forces of attractions between the nuclei and valence electrons of different atoms that join the atoms together.</u></em>
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Which Richter magnitude range can be recorded by instruments but isn't felt? A. less than 2.9 B. 3.0 – 4.9 C. 5.0 – 5.9 D. 6.0 a
algol [13]

A. anything less than 3.0 magnitude on a richters scale usually can't be felt by humans but instruments can pick it up.

3 0
4 years ago
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A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?
emmainna [20.7K]
Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
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8 0
3 years ago
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You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver
Goryan [66]
The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
W= \Delta U = mg \Delta h (1)

where m is the sister's mass, g is the gravitational acceleration and \Delta h is the increase in altitude of the sister with respect to its initial position.

By calling \theta the angle of the chain with respect to the vertical, the increase in altitude is given by
\Delta h = L - L \cos \theta = L(1 - \cos \theta) (2)
where L is the length of the chain.

Putting (2) inside (1), we find
W= m g L (1 - \cos \theta)
from which we can find the mass of the sister:
m =  \frac{W}{g L (1 - \cos \theta)} =  \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg
5 0
3 years ago
52. How did people spend Life in the<br>hunting age-​
kondor19780726 [428]

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7 0
4 years ago
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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
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