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
kaheart [24]
3 years ago
14

A flywheel with radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.400 rad/s2.For a po

int on the rim of the flywheel, what is the magnitude of the tangential acceleration after 2.00 s of acceleration
Physics
1 answer:
oksian1 [2.3K]3 years ago
6 0

Answer:

0.12\ m/s^2

Explanation:

Given that,

The radius of a flywheel, r = 0.3 m

Angular acceleration of a flywheel, \alpha =0.4\ rad/s^2

We need to find the magnitude of the tangential acceleration after 2.00 s of acceleration.

The relation between the tangential and angular acceleration is given by :

a_t=r\alpha \\\\a_t=0.3\times 0.4\\\\a_t=0.12\ m/s^2

So, the required magnitude of tangential acceleration is 0.12\ m/s^2.

You might be interested in
Which of the following is not accurate when desçribing solids?
Kitty [74]
There are no inaccurate items on the list you provided.

And what does 'the following" mean anyway ?
7 0
4 years ago
Which of the following is not an intensive physical property?
Alex Ar [27]

Answer;

-Mass

Explanation;

-An intensive property is a physical quantity whose value does not depend on the amount of the substance for which it is measured. For example, boiling point, density, color, melting point, Odor, temperature, etc.

-Extensive properties on the other hand, do depend on the amount of matter that is present. An extensive property is considered additive for subsystems. Examples of extensive properties include: volume, mass, size, weight. length, etc.


5 0
4 years ago
Read 2 more answers
An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the
Alexxx [7]
(1) The position around equilibrium of an object in simple harmonic motion is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the motion
\omega is the angular frequency.

The velocity is the derivative of the position:
v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)
where 
v_0 = \omega A is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)
where
a_0=\omega^2 A is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
v_0 = \omega A = 4.3 m/s
a_0 = \omega^2 A = 0.65 m/s^2
From the first equation, we get
A= \frac{4.3 }{\omega} (1)
and if we substitute this into the second equation, we find the angular fequency
\omega=0.15 rad/s
while the amplitude is  (using (1)):
A=28.7 m


(b) We found in the previous step that the angular frequency of the motion is
\omega=0.15 rad/s
But the angular frequency is related to the period by
T= \frac{2 \pi}{\omega}
and so, the period is
T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s
5 0
3 years ago
Just helping as usual what about you​
7nadin3 [17]

Answer:

same i just love helping people out

god bless you have a blessed day

Explanation:

6 0
3 years ago
Read 2 more answers
We have all complained that there aren't enough hours in a day. In an attempt to fix that, suppose all the people in the world l
Natali5045456 [20]

Answer:

the duration of a day increases 4.96x10^-11 s

Explanation:

According the exercise:

R=radius of Earth=6.37x10^6 m

mE=mass of Earth=5.97x10^24 kg

m=average mass of people=55 kg

n=number of population=7x10^9

M=total mass=n*m=7x10^9*55=3.85x10^11 kg

v=speed=2.5 m/s

The moment of inertia of population is:

I=(2/3)*M*R^2=(2/3)*3.85x10^11*(6.37x10^6)=1.04x10^25 kg*m^2

The time taken per revolution is:

T=2πR/v=(2π*6.37x10^6)/2.5=1.6x10^7 rev/s

The angular speed is:

w=2π/T=2π/1.6x10^7=3.9x10^-7 rad/s

The angular momentum of population is equal to:

L1=I*w=1.04x10^25*3.9x10^-7=4.08x10^18 kg*m^2/s

The angular momentum of Earth is equal to:

L2=I*w=((2/5)*me*R^2)*(2π/24)=((2/5)*5.97x10^24*(6.37x10^6)^2)*(2π/(24*60*60))=7.1x10^33 kg*m^2/s

The change in length of the day is equal to:

T´=T*(L1/L2)=(24*60*60)*(4.08x10^18/7.1x10^33)=4.96x10^-11 s

8 0
3 years ago
Other questions:
  • The San Andreas fault is an example of which type of tectonic plate boundary?
    8·2 answers
  • Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter
    14·1 answer
  • If two stars are the same size and one is twice the temperature of the other, how much more luminous is the hotter one? quizlit
    14·1 answer
  • Fiberglass, an insulator, can be found in the wals and roofs of some houses and buildings. Why would an insulator be needed insi
    5·2 answers
  • Explain how carbon is cycled between the hydrosphere and geosphere. Use specific examples.
    7·1 answer
  • A student is “accidentally" pushed by Mr. M from the top of NHS. The roof of NHS to the ground (by the
    10·1 answer
  • A bird flaps its wings 8 times per second. what is the frequency of the birds wing flapping?
    14·2 answers
  • Which of these factors is pushing elephant species toward extinction?
    14·2 answers
  • The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32
    8·1 answer
  • Iwill need to use more force to stopa<br> O Lighter mass<br> O Heavier mass
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!