Answer:
2666 kg
0.11567 m/s²
Explanation:
m = Mass of boat
a = Acceleration of boat
From Newton's second law
Force

Force on the first boat is 333.25 N

Hence, mass of the second boat is 2666 kg
Combined mass = 2666+215 = 2881 kg

The acceleration on the combined mass is 0.11567 m/s²
You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.
Answer:
The net torque is 0.0372 N m.
Explanation:
A rotational body with constant angular acceleration satisfies the kinematic equation:
(1)
with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

The negative sign indicates the sphere is slowing down as we expected.
Now with the angular acceleration we can use Newton's second law:
(2)
with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:
With M the mass of the sphere an R its radius, then:

Then (2) is:

Answer:
the molecules have different energy and the system is not in equilibrium
Explanation:
The model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.
So when you book them together the molecules have different energy and the system is not in equilibrium
Answer:
Work done, W = 19.6 J
Explanation:
It is given that,
Mass of the block, m = 5 kg
Speed of the block, v = 10 m/s
The coefficient of kinetic friction between the block and the rough section is 0.2
Distance covered by the block, d = 2 m
As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.


W = 19.6 J
So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.