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Drupady [299]
3 years ago
12

An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 5050 rpm while it is be

ing read, and then is allowed to come to rest over 0.435 s, what is the magnitude of the average angular acceleration of the disk?
Physics
1 answer:
Ber [7]3 years ago
4 0

Answer:

α = 1215.71 rad/s²

Explanation:

The angular acceleration of an object is defined as the time rate of change of angular velocity of he object. The formula for the angular acceleration of an object is given as follows:

α = (ωf - ωi)/Δt

where,

α = angular acceleration of the disk = ?

ωf = Final Angular Velocity = 0 rad/s (Since disk finally stops)

ωi = Initial Angular Velocity = (5050 rpm)(2π rad/rev)(1 min/60 s)

ωi = 528.83 rad/s

Δt = time interval = 0.435 s

Therefore,

α = (0 rad/s - 528.83 rad/s)/(0.435 s)

α = - 1215.71 rad/s²

here, negative sign indicates that the direction of acceleration is opposite to the direction of angular velocity or the angular motion. So, the magnitude of acceleration will be:

<u>α = 1215.71 rad/s²</u>

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The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is
77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

\% Percent = \frac{V_{nucleus}}{V_{atom}}*100

\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100

\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100

\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100

\% Percent = 7.51*10^{-13}*100

\% Percent = 7.51*10^{-11}\%

Therefore the percent of the atom's volume is occupied by mass is 7.51*10^{-11}\%

3 0
2 years ago
Plz help
katrin2010 [14]

The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.

This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.

The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.

4 0
3 years ago
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
3 years ago
A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie
ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, A=7.4\ cm^2=7.4\times 10^{-4}\ m^2

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, B_1=0.5\ T

Final magnetic field, B_2=3\ T

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_2-B_1}{t}

\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}

\epsilon=1.65\times 10^{-3}\ V

Induced current in the loop of wire is given by :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.65\times 10^{-3}}{2.4}

I=6.87\times 10^{-4}\ A

So, the induced current in the loop of wire over this time is 6.87\times 10^{-4}\ A. Hence, this is the required solution.

7 0
3 years ago
I have one but can some one help with the second one?​
alisha [4.7K]

Answer:

it is a, direction

5 0
2 years ago
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