1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
eimsori [14]
3 years ago
8

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

ring that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) what is the instantaneous angular velocity of the disk at the end of the 6.00 s? (d) with the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 6.00 s?
Physics
1 answer:
Klio2033 [76]3 years ago
7 0

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

You might be interested in
A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3
Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

a_c =\dfrac{v^2}{r}

2.32 =\dfrac{3.43^2}{r}

r =\dfrac{3.43^2}{2.32}

   r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 0
3 years ago
A 12.0 V car battery is being used to power the headlights of a car. Each of the two headlights has a power rating of 41.8 Watts
Marina CMI [18]

Answer:

n = 4.35 x 10¹⁹          

Explanation:

Given that

Voltage V= 12 V

Power rating of headlights = 41.8 W

We know that headlights of the car is connected in parallel connection.

We know that

Power = Current x Voltage

P = V I

I=\dfrac{41.8}{12}\ A

I=3.48  A

Therefore the total current will be

I'= 3.48 + 3.48 A

I = 6.96 A

We know that

Charge = Current x time

q= I' t

q= 6.96 x 1

q= 6.96 C

The charge on electron ,e= 1.6 x 10⁻¹⁹

q= n e

n=Number of electron

n=\dfrac{q}{e}

n=\dfrac{6.96}{1.6\times 10^{-19}}

n = 4.35 x 10¹⁹

5 0
3 years ago
A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin
Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain​, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa

The shearing strain is defined as the tangent of the displacement that the block over its length:

SS=tan\theta =\frac{Displacement}{L}  =\frac{10}{40} =0.25

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

Sm=\frac{10208.3333}{0.25} =40833.3332Pa

6 0
3 years ago
If your weight is 100N and you run up a flight of stairs that is 6 m high and it takes
fenix001 [56]

Answer:

power=300watt

Explanation:

5 0
3 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
3 years ago
Other questions:
  • Which nuclear emission has the greatest penetrating power
    13·1 answer
  • A sealed container holds 0.020 moles of ideal nitrogen (n2) gas, at a pressure of 1.5 atmospheres and a temperature of 290k. the
    14·1 answer
  • A postman needs to deliver a parcel to Peter's house 5
    11·1 answer
  • I will mark brainlest! SO HELP ME M8!
    15·1 answer
  • Compare the properties of sodium chloride and sand
    6·1 answer
  • The approximate rotation period of the Moon is: A. 1 day B. 1 week C. I month D. Infinite, since the Moon does not rotate, but k
    6·1 answer
  • Please help me! this is timed!
    15·1 answer
  • In which image below is the angle of refraction the greatest?
    8·1 answer
  • Calculate the average maximum height for all three trials when the mass of the bottle is 0.125 kg, 0.250 kg,
    14·2 answers
  • I will run one mile in under 10 minutes 3 months from today.” is an example of a ___________ goal.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!