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3 years ago

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which statement is true about the atomic mass? a. it is never greater than the atomic number b. it always equal the atomic numbe

r c. it always lowers the atomic number d. it can be greater than the atomic number

2 answers:

Karolina [17]3 years ago

6 0

Answer:

d. it can be greater than the atomic number

Explanation:

Atomic mass number can be defined as the sum of number of protons and number of neutrons present in the nucleus of an atom. On the other hand, atomic number is determined by the number of protons present in the nucleus of the atom. Thus, atomic mass number can be equal or greater than the atomic number but cannot be less than it. Hydrogen has 1 proton and 0 neutrons. So, atomic mass number and atomic number are equal. For rest, the atomic mass number is greater than the atomic number. Thus, option d is correct.

Shtirlitz [24]3 years ago

5 0

D. it can be greater than the atomic number

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

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Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

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3 years ago

An electron is released from rest in a weak electric field given by vector E = -1.30 10-10 N/C Äµ. After the electron has travel

Murljashka [212]

Answer:

v = 6.45 10⁻³ m / s

Explanation:

Electric force is

F = q E

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F- W = m a

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a = 0.228 10² -9.8

a= 13.0 m / s²

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v = 6.45 10⁻³ m / s

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Answer:

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