Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.
Answer:
work done on the object by gravitational force = 0 joules
Explanation:
work = force × displacement
force = mass × acceleration
so,
work = mass × acceleration × displacement
we know that mass= 10 kg , gravitational acceleration= 9.8 
and displacement= 0 m since the object is not moving vertically.
so,
work = 10 × 9.8 × 0 = 0 joules
Answer:255.255.255.224
Explanation:
/20 has a subnet mask of 255.255.240.0
We have 20 on bits for the network and 16 off bits for the host. Since the designer wants 20 to 30 hosts in each subnet we will be having 5 off bits for the host. Therefore we will be having a /27..
Which will be 255.255.255.224
The right answer is A. I'm one lever and two wedges
Hope this helps