Answer:
C. 28.09 amu
Explanation:
The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abundance) and X-30 (29.974 amu, 3.10% abundance).
The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.
The atomic weight is computed as follows:
atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of X-30 × fractional abundance
atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 × 0.0310
atomic weight = 25.8031871 + 1.3531792 + 0.929194
atomic weight = 28.0855603 amu
To 2 decimal place atomic weight = 28.09 amu
Ah ha ! Very interesting question.
Thought-provoking, even.
You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.
Weight = (mass) x (local gravity)
Mass = (weight) / (local gravity)
Mass = (1 Newton) / (local gravity)
"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.
If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:
"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there. And if you don't mind, be quick about it."
The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.
So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth.
That's almost 3.6 pounds of gold, worth over $57,000 !
It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid. Wherever he's willing to go
that has the smallest gravity. That's the place where the largest
mass weighs 1 Newton.
Answer:
A = m³/s³ = [L]³/[T]³ = [L³T⁻³]
B = m³s = [L³T]
Explanation:
We have the equation:
V = At³ + B/t
where, the dimensions of each variable are as follows:
V = m³ = [L]³
t = s = [T]
substituting these in equation, we get:
m³ = A(s)³ + B/s
for the homogeneity of the equation:
A(s)³ = m³
<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>
Also,
B/s = m³
<u>B = m³s = [L³T]</u>
I think u should follow the formulae F=MA. So I think the answer is 120N
