A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
1 answer:
Answer:
-1.43 m/s relative to the shore
Explanation:
Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:
where are the mass of the swimmer and raft, respectively.
are the velocities of the swimmer and the raft after the run, respectively. We can solve for
So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore
You might be interested in
Answer:
k = 40 N/m
Explanation:
A spring's energy is given:
U is the energy in the spring, k is the spring constant and x is the spring displacement.
We are told that the spring stores 5J of energy, therefore, U = 5J. We are also told that the spring is compressed by 0.5m, so the spring x = 0.5m
k = 40 N/m
Hope this helps!
Answer:
a) -1.325 μC
b) 4.17μC
Explanation:
First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:
We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.
The electrostatic force is equal to:
K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2
Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:
Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:
Solving the quadratic equation:
for this values q_1 wil be:
So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC