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
I am not sure what kind of answer you are looking for but the start of friendship.
Answer:
ac = 72 m/s²
Fc = 504 N
Explanation:
We can find the centripetal acceleration of the hammer by using the following formula:
where,
ac = centripetal acceleration = ?
v = constant speed = 12 m/s
r = radius = 2 m
Therefore,
<u>ac = 72 m/s²</u>
<u></u>
Now, the centripetal force applied by the athlete on the hammer will be:
<u>Fc = 504 N</u>
<em>Given that:</em>
mass of the ball (m) = 0.5 Kg ,
ball strikes the wall (v₁) = 5 m/s ,
rebounds in opposite direction (v₂) = 2 m/s,
time duration (t) = 0.01 s,
<em> Determine the force (F) = ?</em>
We know that from Newton's II law,
<em>F = m. a</em> Newtons
(velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>
= m × (v₁ + v₂)/t
= 0.5 × (5 + 2)/0.01
= 350 N
<em>The force acting up on the ball is 350 N</em>
