A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off, it continues to rotate with a constant angular deceleration for 10.0s before coming to rest.Part A

If its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )

Part B

How many revolutions did the centrifuge complete after being turned off?


Answer 1


a_r=389\ rev.s^(-2)

n=58350 rev



time of constant deceleration, t=10\ s


initial angular speed, N_i=3890\ rpm\

Using equation of motion:


0=3890+a_r* 10

a_r=389\ rev.s^(-2)


Using eq. of motion for no. of revolutions, we have:

n=N_i.t+(1)/(2) a_r.t^2

n=3890* 10+0.5* 389* 100

n=58350\ rev

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Wave is defined as a disturbance that travels through a MEDIUM and transfer energy from one point to another without causing any permanent displacement of the medium itself.

The medium through which wave travels varies e.g water, string, air etc

According to the diagram, wave travels through the string held by the man. This string is referred to as the MEDIUM through which the wave moves. The wave generated produces both crest (D) and trough (C) when displaced from its initial position (A)

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When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Starting from rest, it takes two revolutions for the disk to reach full speed. Assuming that the angular acceleration of the disk is constant, what is its angular acceleration while it is speeding up


Final answer:

The angular acceleration of the disk drive in an old computer game system while speeding up is 1256 rad/s². This is calculated using kinematics in rotational motion, given the information on rotations, revolution time, and start from rest.


To calculate the angular acceleration of a disk drive in an old computer game system, we must use the concept of kinematics in rotation. When it is stated that it takes two revolutions to reach full speed, this implies that the total angular displacement is 4π radians (since one full revolution is 2π radians).

Given that the disk drive revolves once every 0.050 seconds, the final angular speed (ω) can be computed as 2π rad/0.050 s = 125.6 rad/s. Since the disk starts from rest, the initial angular speed (ω0) is 0. As a result, the total time taken (t) to reach full speed is 2*0.050s = 0.1 s.

We can then use the equation of motion in rotational form, α = (ω - ω0)/t, to calculate the angular acceleration. Hence the angular acceleration (α) is (125.6 rad/s - 0 rad/s) / 0.1 s = 1256 rad/s². Therefore, the angular acceleration of the disk drive is 1256 rad/s² while it is speeding up.

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Final answer:

The angular acceleration of the disk drive while it is speeding up is 8π rad/s².


The angular acceleration of the disk drive while it is speeding up can be determined by using the formula: angular acceleration = (final angular velocity - initial angular velocity) / time taken. In this case, the initial angular velocity is 0 (since the disk starts from rest) and the final angular velocity is 2 revolutions per 0.050 seconds. To convert revolutions to radians, multiply by 2π. The time taken is the time for two revolutions, so it is 2 * 0.050 seconds. Plugging in these values in the formula, we get:

Angular acceleration = (2 * 2π rad/s - 0) / (2 * 0.050 s) = 8π rad/s²

Learn more about Angular acceleration here:



A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?



The given data is as follows.

      Mass, m = 62 kg,       Initial speed, v_(1) = 6.90 m/s

 Length of rough patch, L = 4.50 m,      coefficient of friction, \mu_(k) = 0.3

 Height of inclined plane, h = 2.50 m

According to energy conservation equation,

     (Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

     K.E_(2) + U_(2) = K.E_(1) + U_(1) - W_(f)

    (1)/(2)mv^(2)_(2) + U_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

    (1)/(2)mv^(2)_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL    

Cancelling the common terms in the above equation, we get

     (1)/(2)v^(2)_(2) = (1)/(2)v^(2)_(1) + gh - \mu_(k)gL

                         = (1)/(2)(6.90)^(2)_(1) + 9.8 m/s^(2) * 2.50 m - (0.3 * 9.8 * 4.50 m)

                         = 36.055 - 13.23

                         = 22.825

               v_(2) = √(2 * 22.825)

                           = 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.



mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s  

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What power lens is needed to correct for farsightednesswherethe uncorrrected near point is 75 cm?


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