Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Speed = distance/time
Speed = 50/5
Speed = 10m/s
Hello.
The answer is:
It creates a spark.
Then thespark can sometimes start a fire or other serious problems.
Have a nice day
Answer:
the mouth I think
Because: the oesophagus is the throat and the mouth is how the food gets to the oesophagus. might be wrong because I'm not sure about the organ part but I'm pretty sure that's all it could be
Answer:
The correct answer is 
Explanation:
The formula for the electron drift speed is given as follows,

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.
Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is
. Converting this number to m³ using very elementary unit conversion we get
. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

if we convert the area from mm³ to m³ we get
.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

which is our final answer.