Aturntable has an angular velocity of 3.5 rad/s. a dust bunny is on the disk of the turntable at a distance of 0.2m from the center of rotation. the coefficient of static friction required to keep the dust bunny from getting slung off is at leasta. 0.51b. 0.45c. 0.33d. 0.12e. 0.25

E. 0.25

Explanation:

Given that

Angular speed ω = 3.5 rad/s

Distance ,r= 0.2 m

Lets take mass of dust bunny = m

We know that

Radial force F = m ω² r

The friction force on the dust bunny Fr

Fr= μ m g

To getting slung off  dust bunny from disk

m ω² r =  μ m g

ω² r =  μ  g

3.5²  x 0.2 =  μ x 10                        ( take g =10 m/s²)

μ = 0.245

μ = 0.25

Therefore answer is E

E. 0.25

your mom

explanation:

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.