As the iron core isprogressively laid down, the mechanism
of iron oxidation changes from a protein dominated
process with H202 being the primary product of
O2 reduction to a mineral surface dominated process
where H20 is the primary product. These results emphasize
the importance of the apoferritin shell in facilitating
iron oxidation in the early stage of iron deposition
prior to significant development of the polynuclear
iron core.
Answer:
when the ball is at its highest in the air.
Explanation:
I don't know for sure, but when the ball is in the air it has potential energy to fall(or something like that).
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
Answer:
Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]
Explanation:
To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.
The gravitational force is equal to:
Fg = (10 * 9.81) = 98.1 [N]
Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.
N - Fg = 0
N = Fg
N = 98.1 [N]
The friction force is defined as the product of normal force by the coefficient of friction.
Ff = N * μ
Ff = 98.1 * 0.4
Ff = 39.24 [N]
By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.
60 - Ff = m * a
60 - 39.24 = 10 * a
a = 2.076[m/^2]
Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N