A computer is reading data from a rotating CD-ROM. At a point that is 0.032 m from the center of the disc, the centripetal acceleration is 111 m/s2. What is the centripetal acceleration at a point that is 0.070 m from the center of the disc?

Question

A computer is reading data from a rotating CD-ROM. At a point that is 0.032 m from the center of the disc, the centripetal acceleration is 111 m/s2. What is the centripetal acceleration at a point that is 0.070 m from the center of the disc?

Answer 1

The entire disc rotates at the same angular velocity w, and centripetal acceleration, rw^2, is proportional to r. So, at 0.070 m from the center, it will be 0.070/0.032 or 2.19 times higher.

Answer 2

For the reaction: C4H10 + O2 �¨ CO2 + H2O ( is this balanced?) , how many moles of CO2 is formed when 5.00 moles C4H10 reacts with O2?

A. 5.00
B. 1.25
C. 20.0

Answer 3

Please use "Post a New Question" for new questions. Your chemistry question is not related to the physics question and answer that preceded it in this thread.

The reaction you wrote is not balanced. After you balance it, the ratio of moles of CO2 (formed) per mole of C4H10 (consumed) will be obvious.

Answer 4

A speed skater goes around a turn that has a radius of 29 m. The skater has a speed of 14 m/s and experiences a centripetal force of 450 N. What is the mass of the skater?

Answer 5

To find the centripetal acceleration at a point that is 0.070 m from the center of the disc, we can use the formula for centripetal acceleration:

Centripetal acceleration (a) = (velocity (v))^2 / radius (r)

Given that the centripetal acceleration at a point that is 0.032 m from the center of the disc is 111 m/s^2, we can find the velocity at that point using the same formula.

Substituting the known values into the formula, we have:

111 m/s^2 = (v)^2 / 0.032 m

To calculate the velocity, we need to rearrange the equation:

(v)^2 = 111 m/s^2 * 0.032 m

(v)^2 = 3.552 m^2/s^2

Now, we can take the square root of both sides to solve for the velocity:

v = √(3.552 m^2/s^2)

v ≈ 1.884 m/s

Now that we know the velocity at the point that is 0.032 m from the center of the disc, we can find the centripetal acceleration at a point that is 0.070 m from the center of the disc.

Using the same formula:

Centripetal acceleration (a) = (velocity (v))^2 / radius (r)

Substituting the values:

a = (1.884 m/s)^2 / 0.070 m

a ≈ 50.913 m/s^2

Therefore, the centripetal acceleration at a point that is 0.070 m from the center of the disc is approximately 50.913 m/s^2.