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e-lub [12.9K]
3 years ago
5

During a hard drive crash the read/write head scrapes against the disk with a coefficient of kinetic friction of µk and normal f

orce of N. Assume that just before the head crash the disk is rotating at ω0 rad/s, and the distance of the head from the disk axis is r. You can ignore any friction at the bearing (rotational axis of the disk). Assume that the disk is uniform, and has radius R and mass M. What is the angular acceleration associated with the torque from the crashed disk head?
Physics
1 answer:
Alex_Xolod [135]3 years ago
3 0

Answer:

 α = \frac{2 \mu \  N}{m \ r}

Explanation:

For this exercise we use Newton's equation for rotational motion

           ∑ τ = I α

the troque is

           α = Fr .r

the moment of inertia of a cylinder is

           I = ½ m r²

we substitute

         fr r = (½ m r²) α

the expression friction is

         fr = μ N

we substitute

         μ N r = ½ m r² α

       

         α = \frac{2 \mu \  N}{m \ r}

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Answer:

True

Explanation:

The atomic mass of oxygen is 16amu, which means the <em>molar mass</em>, the mass of one mole of oxygen atoms (1 mole = 6.02x10²³), is 16g.

5 0
3 years ago
two 2.5 kg balls move away from each other one traveling 3 m/s to the right the other 4 m/s to the left what is the magnitude of
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Answer:

2.5 kg.m/s

Explanation:

Taking left side as positive while right side direction as negative then

Momentum, p= mv where m is the mass of the object and v is the velocity of travel

Momentum for ball moving towards right side=mv=2.5*-3=-7.5 kg.m/s

Momentum for the ball moving towards the left side=mv=2.5*4=10 kg.m/s

Total momentum=-7.5 kg.m/s+10 kg.m/s=2.5 kg.m/s

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3 years ago
What are the order of the 7 prefixes you are required to know in order from largest to smallest
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10^9 giga, 10^6 mega, 10^3 kilo, 10^-3 milli, 10^-6 micro, 10^-9 nano, 10^-12 pico
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If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat
Maksim231197 [3]

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

    v_{max} = Aw

  Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to w\\ of the oscillation .

  Since  w = 2 \pi f

      v_{max}^{|}/v_{max} = 2f/f = 2

  Where v_{max}^{|} is the maximum speed when frequency is doubled .

6 0
3 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
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