Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
Answer:
Centripetal Acceleration = 88826.44 m/s²
RCF = 9054.7
Explanation:
First, we will find the value of the centripetal acceleration by using the following formula:
where,
v = linear speed of liquid or separator = rω
ω = angular speed of liquid or separator = (9000 rpm)(2π rad/rev)(1 min/60 s) = 942.48 rad/s
r = radius of seperator = diameter/2 = 20 cm/2 = 10 cm = 0.1 m
Therefore,
<u>Centripetal Acceleration = 88826.44 m/s²</u>
Now, for the RCF:
<u>RCF = 9054.7</u>