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mestny [16]
3 years ago
15

How would the composition of an atom change if both the atomic number and mass number each increase by one? A) The atom would ha

ve 1 more neutron. B) The atom would have one more proton and electron. C) The atom would have one more proton and one more neutron. D) The atom would have one more proton, one more neutron and one more electron.
Physics
1 answer:
weqwewe [10]3 years ago
8 0
Choice-B is the correct one. 

-- The atomic number is the number of protons in the nucleus.
-- Each proton in the nucleus is usually matched by one electron in the 'cloud'.
-- The addition of a proton OR a neutron increases the mass number by 1 .
-- Electrons have such small mass that they don't figure into the atomic mass at all.
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What is the answer?
Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE    =   KE

mgh = 1/2 mv^2     divide both sides of the equation by 'm'

gh    = 1/2 v^2             multiply both sides by 2

2 gh = v^2               take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

3 0
2 years ago
A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0
3 years ago
What are three types of variables in a controlled experiments?
blsea [12.9K]

The answers to your question are,

Independent, Dependent, and Control.

-Mabel <3

7 0
3 years ago
What is your opinion on Fluorinated gases
Nikolay [14]

Answer: Fluorinated gases (F-gases) are man-made gases that can stay in the atmosphere for centuries and contribute to a global greenhouse effect. this is extremely bad for the earth and is causing downpour on our plants and animals. Why would we creat gases we can not get rid of its ver diapointing

3 0
3 years ago
A ball has a mass of 0.1 kg and an initial velocity of 20 m/s. The ball is given an acceleration of 30 m/s2 for 5 s. What is the
djverab [1.8K]

Answer:

acceleration = v-u /t

30- 20/5

= 10/5 = 2m/sec²

Force = mass * acceleration

Force = 0.1 * 2

Force = 0.2 Newton

6 0
3 years ago
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