You have to make an assumption about how the racecar is rolling. Normally, our cars roll without slipping, meaning that the wheels do not slip against the ground. In other words, one revolution of the wheels corresponds to 2πR distance covered on the ground.

Another way of writing the no-slip condition is v_t = Rω, where v_t is the speed of the car. Try using this relationship to find the energies.

The Earth is rotating because it has angular momentum. When an event such as a bridge collapse occurs, the distribution of mass on Earth changes, but the total angular momentum is conserved.

Consider the equation L = I ω. The angular momentum L is conserved, but the moment of inertia I has changed. We know that moment of inertia is greatest when the mass is distributed closer to the axis of rotation.If an underwater bridge collapses and falls, does this increase or decrease the moment of inertia of the Earth?

Once you know whether the moment of inertia increases or decreases, you can determine from the equation whether the angular speed ω has to increase or decrease to keep L constant. How does the angular speed of Earth affect the length of the day?

This problem is about buoyancy. The buoyant force is equal to the weight of the displaced fluid. In other words, put an object in a fluid and it will displace fluid, causing the level to rise. If an object floats, that means the buoyant force is enough to cancel out the force of gravity on that object. If it sinks, the object isn't able to displace enough fluid to cancel out gravity.

For this problem, it will be helpful to draw free body diagrams in the before and after case. Before, you have the boat+ring combination floating and two forces acting on it: gravity and buoyancy. Remember that the force of gravity is for the combined masses, boat and ring. After, you have the boat still floating, but the ring on the bottom of the glass. Like we noted earlier, the ring sinks because the buoyant force is not enough to cancel out gravity. In this case the normal force from the glass does the rest of the job.

Now let's start writing some equations. In the Before case, we have the force of gravity equal to the buoyant force:

ρ * V_both * g = (m_boat + m_ring) * g

where V_both is the volume of fluid displaced by the boat with the ring in it.

In the After case, we have a similar equation for the boat:

ρ * V_boat * g = m_boat * g

and we can write an inequality for the ring because we know the buoyant force is weaker than gravity:

ρ * V_ring * g < m_ring * g

Now let's try to solve for something useful. Adding the last two expressions gives another inequality:

ρ * (V_boat + V_ring) * g < (m_boat + m_ring) * g

and now we can plug in the first equation:

ρ * (V_boat + V_ring) * g < ρ * V_both * g

and arrive at the conclusion:

V_boat + V_ring < V_both

in other words, the volume of fluid displaced by the boat with the ring in it is greater than the total volume displaced when the objects are separate. So when the ring sinks, less fluid is being displaced so the water level lowers.

This problem is quite simple if you have a good diagram. Draw this scenario but instead actually drawing an arm, just draw a rod with force arrows on it. From here, it should be obvious that this is a simple equilibrium question where the torques (F * d) sum to zero.

Remember that the force of friction is always going against the motion of the object. In this case, gravity is making the child go downward, therefore you should consider that the force of friction up the slide. This means, it would be pulling the child back unfortunately.

The force of friction may be in the direction opposite to the motion of the tablecloth

If you ever forget this just remember that

W = F • Δx

therefore if the force is going in the opposite direction then it is negative work.

To answer your first question, you need to set F = k*s (F=force,k=spring constant,s=stretch from equilibrium) and solve for the spring constant, if you are looking for that initially.

For this problem, you need the kinematics equation for falling objects (x = (at^2)/2 + v_i*t + x_i). Also, you need the time (t) it will take for the bully to be under the window. Therefore, you just need to find t in v_bully = d/t (d = distance of the bully from under the window and v_bully = bully's constant speed). Then solve for v_i in the kinematics equations.

Keep in mind that x_i, a, and x are the initial height, acceleration of gravity (a<0), and height which you want to hit the bully at.

If you keep in mind that w^2 = k/m and w = 2pi*f (f is frequency, m is total mass), then you can solve for the total mass m in terms of f, w, and some constants. However, do not forget to subtract the mass of the chamber from m to get the mass of the astronaut.

For the first question, you can simply think of the problem as a balance of forces. You need to balance them by equating the buoyant force and the force of gravity of some arbitrary mass (call it M). After solving for M, just subtract off the mass of the raft to give you the total mass you can add without the raft sinking.

For the second question, are you asking how the weight change if you evaluated the problem as a perfect rectangular box instead of a box with rounded edges? If so, then it would change how strong the buoyant force is because the volume of the raft would be different.

V = L*D^2*8 vs V = L*D*7 + pi*(D/2)^2, where

L = length of logs

D = diameter of logs