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Sunny_sXe [5.5K]
2 years ago
8

A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 re

volutions before coming to rest.
Find the constant angular acceleration (in rad/s2) of the centrifuge.


______rad/s2

Physics
1 answer:
eduard2 years ago
7 0

The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

<h3> Constant angular acceleration</h3>

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

  • ωf is the final angular velocity when the centrifuge stops = 0
  • ωi is the initial angular velocity
  • θ is angular displacement
  • α is angular acceleration

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

#SPJ1

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1 km/h = 5/18 m/s

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When an object moves, where does the energy come from?
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Ans: Kinetic and potential energies are found in all objects. If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick.
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How much force is needed to accelerate a 66 kg skier 1 m/sec/sec?
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66 N because F=ma. m is in kg and a is in m/s^2 so you can just multiply them
4 0
3 years ago
Ball 1, with a mass of 100 g and traveling at 10 m/s, collides head-on with ball 2, which has a mass of 300 g and is initially a
stiks02 [169]

Answer:

Perfectly Elastic (after collision):

Velocity of Ball 1 = v₁ = -5m/s

Velocity of Ball 2 = v₂ = 5m/s

In the case of a collision, assuming perfectly elasticity, ball 2 would move at a speed of 5m/s and ball 1 would switch direction moving at a now lower speed of 5m/s.

Perfectly Inelastic (after collision):

Velocity of Ball 1 = v₁ = 0m/s

Velocity of Ball 2 = v₂ = 0m/s

Explanation:

Perfectly Elastic:

No kinetic energy is lost via transfer into any other kind, e.g. heat, so we just use the conservation of momentum formula:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.1)(10) + (0.3)(0) = (0.1)v₁ + (0.3)v₂  (×10)

v₁ + 3v₂ = 10 → v₁ = 10 - 3v₂

Now, we have to be a bit innovative and use the kinetic energy formula for both balls:

KE = ¹/₂mv²

Firstly, before the collision:

Ball 1:

(KE)₁ = ¹/₂(0.1)(10)²

(KE)₁ = 5

Ball 2:

(KE)₂ = 0 ← it was at rest

Now after the collision:

Ball 1:

(KE)₁ = ¹/₂(0.1)(v₁)²

(KE)₁ = ¹/₂₀(v₁)²

Ball 2:

(KE)₂ = ¹/₂(0.3)(v₂)²

(KE)₂ = ³/₂₀(v₂)²

Since we are considering a perfectly elastic collision, no energy is lost in the collision so the sum of the kinetic energy of the two balls before and after the collisons has to be equal, hence we can formulate the following:

5 + 0 = ¹/₂₀(v₁)² + ³/₂₀(v₂)²

100 = (v₁)² + 3(v₂)²

Now, we can plug in the equation we found for v₁ from earlier and solve:

100 = (10 - 3(v₂))² + 3(v₂)²

100 = 100 - 60(v₂) + 9(v₂)² + 3(v₂)²

12(v₂)² - 60(v₂) = 0

(v₂)² - 5(v₂) = 0

(v₂)((v₂) - 5) = 0

v₂ = 0 OR v₂ - 5 = 0 → v₂ = 5

v₁ = 10 OR v₁ = 10 - 3(5) = -5 m/s

Perfectly Inelastic:

Final velocities would be 0;

All kinetic energy would be transferred in the collision, the balls would join together to become one in such a case.

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inserting the values

ΔP = (0.5) (1000) ((0.28)² - (0.14)² )

ΔP = 29.4 Pa

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3 years ago
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