What is the average acceleration of the particle between 0 seconds and 4 seconds?
1 answer:
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Answer:
Distance covered is 37.63 ft
Solution:
As per the question:
Circumference of the bowling ball, C = 27 in
Weight of the bowling ball, w = 14.8 lb
Radius of gyration, k = 3.43 in
Velocity of release of the ball, v' = 23 ft/s = 276 in/s
Coefficient of friction,
Now,
We know that the circumference of the circle is given by:
C =
27 =
Also, in case of rolling, we know that:
Angular velocity,
Now, applying the conservation of angular momentum along the floor:
Initial angular momentum = Final angular momentum
Moment of Inertia, I =
v = 99.56 in/s
Now,
Friction force, f =
Also, acceleration of the ball can be computed as:
Now, the distance, d covered by the ball before rolling without slipping:
d = 451.65 in = 37.53 ft
Answer:
<em>0.228 nm</em>
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Explanation:
Atomic mass number of copper = 64
but an atomic mass unit = 1.66 x 10^-27 kg
therefore, the mass of the copper atom m = 64 x 1.66 x 10^-27 kg = 1.06 x 10^-25 kg
The number of atoms in this mass n = ρ/m
where ρ is the density of copper = 8.96 x 10^3 kg/m^3
==> n = (8.96 x 10^3)/(1.06 x 10^-25) = 8.45 x 10^28 atoms/m^3
We know that the volume occupied by this amount of atoms n =
where a is the lattice constant
equating, we have
8.45 x 10^28 =
a = 4.389 x 10^9
we also know that
d = 1/a
where d is the smallest distance between the two copper atom.
d = 1/(4.389 x 10^9) = 2.28 x 10^-10 m
==> <em>0.228 nm</em>