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Whitepunk [10]
3 years ago
8

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplishe

d 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?
Physics
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

a_r=389\ rev.s^{-2}

n=58350 rev

Explanation:

Given:

time of constant deceleration, t=10\ s

A.

initial angular speed, N_i=3890\ rpm\

<u>Using equation of motion:</u>

N_f=N_i+a_r.t

0=3890+a_r\times 10

a_r=389\ rev.s^{-2}

B.

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

n=N_i.t+\frac{1}{2} a_r.t^2

n=3890\times 10+0.5\times 389\times 100

n=58350\ rev

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3 years ago
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

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h2 = hf + x2(hfg)

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6 0
3 years ago
A car's bumper is designed to withstand a 6.84 km/h (1.9-m/s) collision with an immovable object without damage to the body of t
den301095 [7]

Answer:

F = 5.256 x 10^{3} N

Explanation:

From the work energy theorem we know that:

The net work done on a particle equals the change in the particles kinetic energy:

W = F.d, ΔK =\frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i}  , F.d = \frac{1}{2}mv^{2}_{f} -\frac{1}{2} mv^{2}_{i}

where:

W = work done by the force

F = Force

d = Distance travelled

m = Mass of the car

vf, vi = final and initial velocity of the car

kf, ki = final and initial kinetic energy of the car

Given the parameters;

m = 830kg

vi = 1.9 m/s

vf = 0 km/h

d = 0.285 m

Inserting the information we have:

F.d = \frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i}

F = \frac{\frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i} }{d}

F = \frac{ 0   - \frac{1}{2}  X830 X 1.9^{2} }{0.285}

F = 5.256 x 10^{3} N

3 0
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