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:
1362000 kgm/s
Explanation:
So the total mass combination of the plane and the people inside it is
M = 35000 + 160*65 = 45400 kg
After 15 seconds at an acceleration of 2 m/s2, the plane speed would be
V = 2*15 = 30 m/s
So the magnitude of the plane 15s after brakes are released is
MV = 45400 * 30 = 1362000 kgm/s
Answer:
60words/minute
Explanation:
If Sunitha can type 1800 words in half an hour, this can be expressed as;
1800 words = 30 minutes
To get her typing speed per minute, we will use the formula
Speed = Number of words/Time used
Typing speed = 1800/30
Typing speed = 60words/minute
Hence her typing speed in words per minute is 60words/minute
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
