Heya!!
For calculate aceleration, lets applicate second law of Newton:
<u>Δ Being Δ</u>
F = Force = 4 N
m = mass = 36 kg
a = Aceleration = ?
⇒ Let's replace according the formula:
⇒ Clear acceleration and resolve it:
Result:
The aceleration is of <u>0,111 meters per second squared (m/s²)</u>
Good Luck!!
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Answer:
x = -1.20 m
y = -1.12 m
Explanation:
as we know that four masses and their position is given as
5.0 kg (0, 0)
2.9 kg (0, 3.2)
4 kg (2.5, 0)
8.3 kg (x, y)
As we know that the formula of center of gravity is given as
Similarly for y direction we have
In zero order reactions the rate of reaction is independent of reactant concentrations. That is the rate of the reaction does not vary with increasing nor decreasing reactants concentrations. On the other hand, first order reaction are reactions in which the rate of reaction is directly proportional to the concentration of the reacting substance (reactants). In this case, i believe the rate of reaction will triple( increase by a factor of 3)
