Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
The new acceleration is 
Explanation:
We can answer this problem by applying Newton's second law, which states that:

where
F is the net force on an object
m is the mass of the object
a is its acceleration
The equation can be rewritten as

In this problem, the initial acceleration is

Later:
- The net force is tripled: 
- The mass is halved: 
Therefore, the new acceleration is:

which means that the new acceleration is 6 times the original acceleration, therefore

Learn more about acceleration:
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Answer:
F = M a
W = M g equivalent equation to express weight of object of mass M
M = W / g = 2867 N / 9.8 m/s^2 = 292.6 kg