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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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Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
3 years ago
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Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s
frez [133]

Answer:

The Heavier Firefighter

Explanation:

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2 years ago
You and a friend frequently play a trombone duet in a jazz band. During such performances it is critical that the two instrument
Rashid [163]

Answer:

f₂ = 468.67 Hz

Explanation:

A beat is a sudden increase and decrease of sound. The beats are produced through the interference of two sound waves of slightly different frequencies. Now we have the following data:

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Beat Frequency = fb

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fb = n/t

fb = 4 beats/ 3 s

fb = 1.33 Hz

Another formula for beat frequency is:

fb = f₁ - f₂

f₂ =  f₁ - fb

f₂ = 470 Hz - 1.33 Hz

<u>f₂ = 468.67 Hz</u>

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Answer:

i think the anwer is C

Explanation:

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