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3 years ago

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A CD player rotates at a variable speed so that a laser can scan pits and lands on the disk’s bottom surface at a constant tange

ntial speed of 1.2 m/s. The disk has a moment of inertia of 1.2 x 10-4 kg m2and the music is first detected when the laser is located 15 mm from the disk’s center. Assuming the disk started from rest, find the work done by the motor during this start-up.

1 answer:

LekaFEV [45]3 years ago

6 0

Answer:

W = 0.384 J

Explanation:

Work and energy in the rotational movement are related

W = ΔK = $K_{f}$ - K₀

W = ½ I $w_{f}$ ² - 1 /2 I w₀²

where W isthe work, I is the moment of inertia and w angular velocity

With the system part of the rest the initial angular speed is zero (w₀ = 0)

The angular and linear quantities are related

v = w r

w = v / r

Let's replace

W = ½ I (v / r)²

Let's calculate

W = ½ 1.2 10⁻⁴ 1.2² / (1.5 10⁻²)²

W = 0.384 J

W = 38.4 J

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Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangmen

murzikaleks [220]

Answer:

a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,

d) W = - 4.93 10⁻²⁸ J

Explanation:

a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m

If we use the law of conservation of energy, work is the change in energy of the system

W = ΔU = U_∞ -U

the potential energy for point charges is

U =k $\sum \frac{q_i q_j}{r_{ij} }$

in this case we only have two particles

U = k $\frac{q_1q_2}{r_{12} }$

the distance is

r₁₂ = $\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }$

r₁₂ = $\sqrt{ 0 + ( 2-0)^2}$ Ra 0 + (2-0)

r₁₂ = √2= 1.4142 m

we substitute

W = k \sum \frac{q_i q_j}{r_{ij} }

let's calculate

W = $\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}$ 9 109 1.6 10-19 1.6 10-19 / 1.4142

W = 1.63 10⁻²⁸ J

b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0

in this case we have two fixed electrons

U = k $( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )$

in this case all charges are electrons

q₁ = q₂ = q₃ = q

W = U = k q² $( \frac{1}{r_{13} } + \frac{1}{r_{23} } )$

the distances are

r₁₃ = $\sqrt{(3-0)^2 + 0}$ RA (3.00 -0) 2 + 0

r₁₃ = 3

r₂₃ = $\sqrt{ 3^2 + 2^2}$ Ra (3 0) 2 + (2 0) 2

r₂₃ = √13

r₂₃ = 3.606 m

let's look for the job

W = U

let's calculate

W = ${9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )$

W = 1.407 10⁻²⁷ J

c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,

y₄ = 4.00 m

W = U = k $( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )$

all charges are equal q₁ = q₂ = q₃ = q₄ = q

W = k q² $(\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )$

let's look for the distances

r₁₄ = $\sqrt{3^2 +4^2}$

r₁₄ = 5 m

r₂₄ = $\sqrt{3^2 + ( 4-2)^2}$

r₂₄ = √13 = 3.606 m

r₃₄ = $\sqrt{(3-3)^2 + (4-0)^2}$

r₃₄ = 4 m

we calculate

W = 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )$

W = 1.68 10⁻²⁸ J

d) we take the proton to the location x5 = 1m y5 = 1m

W = U = k $( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )$

in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value

W = - k q² $( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )$

we look for distances

r₁₅ = $\sqrt{ 1^2 +1^2}$ Ra (1-0) 2 + (1-0) 2

r₁₅ = √ 2 = 1.4142 m

r₂₅ = $\sqrt{ (2-1)^2 +1^2}$

r₂₅ = √2 = 1.4142 m

r₃₅ = $\sqrt{ ( 3-1)^2 +1^2}$

r₃₅ = √5 = 2.236 m

r₄₅ = $\sqrt{ (3-1)^2 + (4-1)^2}$

r₄₅ = √13 = 3.606 m

we calculate

W = - 9 10⁹ (1.6 10⁻¹⁹)² $( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )$

W = - 4.93 10⁻²⁸ J

3 0

3 years ago

Using the rules for significant figures, what do you get when you add 24.545 and 307.3?

lions [1.4K]

When you add 24.545+307.3= 331.845

You will go with the number with the smallest number, which is 307.3.

So answer will be 331.8

3 0

4 years ago

Read 2 more answers

If a stationary object experiences a downward force of 12 n and an upward force of 12 n, what is the motion of the object?

mojhsa [17]

The motion would not move at all, because the forces that are acting on it are equal and are acting in opposite directions

3 0

3 years ago

3 Below, someone is trying to balance a plank with

belka [17]

Answer:

a. The moment of the 4 N force is 16 N·m clockwise

b. The moment of the 6 N force is 12 N·m anticlockwise

Explanation:

In the figure, we have;

The distance from the point 'O', to the 6 N force = 2 m

The position of the 6 N force relative to the point 'O' = To the left of 'O'

The distance from the point 'O', to the 4 N force = 4 m

The position of the 4 N force relative to the point 'O' = To the right of 'O'

a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point

a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise

b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.

8 0

3 years ago

Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 1

DIA [1.3K]

Explanation:

1. Distance, d = 5 m

Time, t = 10 s

Speed = distance/time

$v=\dfrac{5}{10}=0.5\ m/s$

2. Mass, m = 10 kg

Acceleration, a = 3 m/s³

Force, F = mass (m) × acceleration (a)

F = 10 × 3

= 20 N

3. Mass, m = 7 kg

Height, h = 4 m

Potential energy, E = mgh

E = 7 × 9.8 × 4

E = 274.4 J

Hence, this is the required solution.

6 0

3 years ago