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Angelina_Jolie [31]
2 years ago
13

A compact disc (CD) is played by a cd player, which uses a laser to read the tracks on the disc. The disc spins initially at app

roximately 200 rotations per minute (rpm), and increases to a maximum of approximately 500 rpm as the laser spirals inward towards the center of the disc. This ensures that the laser covers equal distance in an equal amount of time during playback. During the time that the cd is played, which of the following statements is true?
a. The laser tracking mechanism experiences a changing tangential velocity
b. The laser tracking mechanism experiences a constant angular velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration
Physics
1 answer:
uranmaximum [27]2 years ago
4 0

Answer:

a. The laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Explanation:

a. The laser tracking mechanism experiences a changing tangential velocity

This is because the tangential velocity v = rω where r = radius of disc and ω  = angular speed of discs. Since r is constant, v ∝ ω.

Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.

So, the laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.

Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.

So, The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.

Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.

So, the laser tracking mechanism experiences a non-zero tangential acceleration

Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
pshichka [43]

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A. 4,9 m/s2

B. 2,0 m/s2

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In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

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\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

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3 years ago
An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi
BARSIC [14]

Answer:

B= 55.6×10^(-7) Tesla

Explanation:

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r: distance from the wire

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B= (4π×10^(-7)×125)/(2π×4.5)

B= 55.6×10^(-7) Tesla

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