Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal 6.00×105 g (where g is the acceleration due to gravity).

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² =  5880000 / 0.018

ω ² =  326666667

ω = 18074 rad/sec

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.