A 62.0 kg water skier at rest jumps from the dock into a 775 kg boat at rest on the east side of the dock. If the velocity of the skier is 4.50 m/s as she leaves the dock, what is the final velocity of the skier and boat? 2.78 m/s to the east
2.78 m/s to the west
0.360 to the west
0.360 to the east

The final velocity of the skier and boat is 0.33 m/s to the east.

Explanation:

We can find the final velocity of the skier by conservation of linear momentum:

Where:  

: is the mass of the water skier = 62.0 kg

: is the mass of the boat = 775 kg

: is the initial velocity of the skier = 4.50 m/s (as she leaves the dock)

: is the initial velocity of the boat = 0 (it is at rest)

: is the final velocity of the skier =?

: is the final velocity of the boat =?

Since the final velocity of the skier is the same that the velocity of the boat () we have:

[tex]v_{f} = \frac{m_{s}v_{s_{i}}}{m_{s} + m_{b}} = \frac{62.0 kg*4.50 m/s}{62.0 kg + 775 kg} = 0.33 m/s                  

Therefore, the final velocity of the skier and boat is 0.33 m/s to the east.

I hope it helps you!                                                          

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14