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Mice21 [21]
3 years ago
11

When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Star

ting from rest, it takes two revolutions for the disk to reach full speed. Assuming that the angular acceleration of the disk is constant, what is its angular acceleration while it is speeding up
Physics
1 answer:
Gemiola [76]3 years ago
3 0

Answer:

α = 200*π rad/sec²

Explanation:

  • If the disk drive revolves once every 0.05 sec, this means that the angular velocity can be expressed as follows:

        \omega = \frac{2*\pi}{0.05s} = 40*\pi rad/sec

  • (just applying the definition of angular velocity)
  • Now, it the disk started from rest, and took two revolutions to reach to full speed, assuming that the angular acceleration is constant, we can find the angular acceleration applying this kinematic equation:

        \omega_{f} ^{2} -\omega_{0} ^{2} = 2*\alpha * \Delta\theta

  • where Δθ, is the angle rotated while it went from rest to full speed, i.e., 2 revolutions, or 2*π rads.
  • Replacing by the value that we found for ωf (as ω₀ = 0), we  can solve for α :

       \alpha =\frac{\omega_{f}^{2} }{2*\Delta\theta} = \frac{(40*\pi)^{2} }{2*2*2*\pi } =\\ \\ \alpha = 200*\pi rad/sec2

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A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0
2 years ago
Which are characteristics of electromagnetic waves?check all that apply.
emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces  --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light  --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, c=3\cdot 10^8 m/s)

travel only through matter  --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space  --> FALSE, electromagnetic waves can also travel through matter

can bend around objects  --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other  --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave

8 0
3 years ago
Read 2 more answers
A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tensio
BARSIC [14]

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

<h3>At the bottom condition</h3>

T - m g = m a

T =  m ω² r  + m g

<h3>At the top condition</h3>

T + m g = m a

T=  m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

8 0
3 years ago
Rahman drops two stones P and Q each of mass 1 kg and 2 kg simultaneously from
kolbaska11 [484]

This question involves the use of the equations of motion for vertical motion.

The time taken by the stones P and Q to reach the ground is the same, that is "2 s".

The velocity with which Q hits the ground is "20 m/s".

The time taken by the stones to reach the ground can be calculated by using the second equation of motion for the vertical motion:

h = v_it+\frac{1}{2}gt^2

For both the stones P and Q:

h = height = 20 m

v_i = initial velocity = 0 m/s

t = time = ?

g = acceleration due to gravity = 10 m/s²

Therefore,

20\ m = (0\ m/s)t+\frac{1}{2}(10\ m/s^2)(t)^2\\\\t = \sqrt{\frac{20\ m}{5\ m/s^2}}

<u>t = 2 s</u>

<u></u>

Hence, the time taken by both the stones to reach the ground <u>is the same</u>.

To find the final velocity of stone Q we will use the first equation of motion for the vertical motion:

v_f=v_i+gt\\v_f = 0\ m/s+(10\ m/s^2)(2\ s)\\v_f = 20\ m/s

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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c the galaxy because on a nice night you cant see the milkly way

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