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
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Explanation:
When pH of the solution is 11.
..(1)
At pH = 11, the concentration of ions is .
When the pH of the solution is 6.
..(2)
At pH = 6, the concentration of ions is .
On dividing (1) by (2).
The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is .
Difference between the ions at both pH:
This means that Hydrogen ions in a solution at pH = 7 has ions fewer than in a solution at a pH = 6
S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
Find t
15 = 1/2(3.2)t^2
15 = 3.2t^2/2
30 = 3.2t^2
30/ 3.2 = 9.38
Square root of 9.38 = 3.06
It takes 3.06 seconds
Answer:
331.7m/s
Explanation:
Given parameters:
Initial velocity = 100m/s
Acceleration = 50m/s²
Distance = 1km = 1000m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to apply the right motion equation shown below;
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Now insert the parameters and solve;
v² = 100² + (2 x 50 x 1000)
v² = 110000
v = √110000 = 331.7m/s