Answer:
P= 454.11 N
Explanation:
Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
Answer:
The equation for the object's displacement is 
Explanation:
Given:
m = 16 lb
δ = 3 in
The stiffness is:

The angular speed is:

The damping force is:

Where
FD = 20 lb
u = 4 ft/s = 48 in/s
Replacing:

The critical damping is equal:

Like cc>c the system is undamped
The equilibrium expression is:

Via half-life equation we have:

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

There is 6.25 grams left of Ra-229 after 12 minutes.
Answer:
The nature of volcanic eruptions is highly dependent on magma viscosity and also on dissolved gas content. ... long it takes the treacle to flow from one end of a boiling tube to the other.