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2 years ago

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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?

1 answer:

Alex_Xolod [135]2 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:

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c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

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b) The increase in velocity shows that there has been acceleration.

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= $\frac{40}{2}$

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3 years ago

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Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

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$v_x = \frac{67}{4.5}$

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Here,

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$-1.23= 4.5v_y - 99.225$

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The direction of the velocity will be given by the tangent of the components, then

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$\theta = tan^{-1} (\frac{21.77}{14.89})$

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The magnitude is given vectorially as,

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$|V| = \sqrt{14.89^2 +21.77^2}$

$|V| = 26.37m/s$

Therefore the angle is 55.59° and the velocity is 26.37m/s

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2 years ago

Which is an example of gaining a static charge by conduction?

lianna [129]

Answer:

shuffling shoes

Explanation:

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3 years ago

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

$E_{m} = E_{k} + E_{p}$

$E_{m} = \frac{1}{2}v^{2} + gh$

Where:

Ek is the kinetic energy

Ep is the potential energy

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g is the gravity = 9.81 m/s²

h is the height = 58 m

$E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg$

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

$P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW$

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

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2 years ago

I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)

Helga [31]

Hello! I can help you with this!

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If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.

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2 years ago