I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps
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.
Answer:
Explanation:
Let the volume below water be v . Then
buoyant force = v d g where d is density of water , g is acceleration due to gravity
= v x 1000 x g
weight of wood piece = volume x density of wood x g
= .6 x 600 x g
for equilibrium while floating
buoyant force = weight
= v x 1000 x g = .6 x 600 x g
v = .36 m²
volume above water or volume exposed = .6 - .36
= .24 m²
When immersed completely ,
buoyant force = .6 x 1000 x 9.8
= 5880 N
weight of wood
= .6 x 600 x g
= 3528 N
buoyant force is more than the weight . In order to equalise them for floating with full volume in water
weight required = 5880 - 3528
= 2352 N.
The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)
Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation
v = vo + at
When you plug in the values you get
v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.
