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7nadin3 [17]
10 months ago
6

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

e they switch the mirror from rotating clockwise at 115 rad/s to counter clockwise at 115 rad/s over the course of 85 seconds. Assume constant angular acceleration.
QUESTION: how many revolutions does the rotating mirror go through when it gets to a 0 angular velocity along the way to new speed. SHOW WORKING OUT
Physics
1 answer:
iris [78.8K]10 months ago
7 0

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115  - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

#SPJ1

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A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e
alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

K.E=\frac{1mv^2}{2}

K.E= kinetic energy = 0.991\times 10^{-19}J

m= mass of an electron = 9.109\times 10^{-31}kg

v= velocity of object = ?

Putting in the values in the equation:

0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}

v=466462m/s

The speed necessary for the electron to have this energy is 466462 m/s

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A massive tractor rolls down a country road. in a perfectly inelastic collision, a small sports car runs into the machine from b
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The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb
topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

F = \frac{mv^2}{r}

The relation between linear velocity and the angular velocity is

v = \omega r

So,

F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r  =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}

As can be seen, the maximum velocity for the projectile without breaking the cable is \sqrt{1500r}, where r is the length of the cable.

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3 years ago
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Answer:

Explanation:

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F in this case is the gravity acting on the 2kg object. Acceleration of gravity is 9.8 m/s^2. SF=ma, so

F = 2kg*(9.8 m/s^2) = 19.6 N

Now use this force to determine the mass of the object on the table:

F=ma

19.6 N (1N=kg*m/s^2) = m*(1.8 m/s^2)

m = 10.89 kg

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