A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 times before coming to rest. Find the constant angular acceleration of the centrifuge (in rad/s2). Consider the direction of the initial angular velocity to be the positive direction, and include the appropriate sign in your result.

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

Explanation:

Given:

Initial angular velocity, wi = 3650 rev/min

Converting from rev/min to rad/sec,

3650 rev/min × 2pi rad/1 rev × 1 min/60 secs

= 382.23 rad/s

Final angular velocity, wf = 0 rad/s (comes to rest)

Angular displacement, Ѳ = 46 times

= 46 revs

= 46 rev × 2pi rad/1 rev

= 289.03 rad

Using equation of angular motion,

wf^2 = wi^2 + 2 × ao × Ѳ

Where,

ao = angular acceleration

0 = 382.23^2 + 2 × 289.03 × ao

Solving for ao,

ao = -252.75 rad/s^2

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.