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
svet-max [94.6K]
3 years ago
6

Determine the total moment of inertia of a merry-go round with 5 children sitting on it. Of the five children, four are seated a

t the edge of the merry-go-round, and the fifth child is sitting exactly at the center. The children have the same mass of 30.0 kg. You may consider the merry-go-round as a cylinder of mass 150 kg and a radius 2.0 meters.
Physics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

Explanation:

Given that,

We have five children.

Each of mass m =30kg

They sit on a merry go round

Mass of Merry go round M= 150kg

Radius of Merry go round is r =2m

Four children sit at the edge of the merry go round but one child sit at the centre.

The four child that sit at the edge are 2m from the centre of the merry go round but the one at the centre is 0m from the centre

Moment of inertia?

Moment of inertia is given as

I=Σmi•ri²

For the question, the moment of inertia is the combination of inertial of child and the merry go round

I= I(merry go round) + I(four child)+ I(last child)

The merry go round is assumed to be a solid cylinder, so it is going to have the moment of inertia of solid cylinder

Then,

I(merry go round ) =½ Mr²

Also, Four of the child has the same moment of inertia, they are 2m form the centre of the merry go round why the last child has no moment of inertia

I= I(merry go round) + I(four child) +I(last child )

I= ½Mr² + 4mr² + mr'²

I = ½ × 150 ×2² + 4×30×2² + 30×0²

I = 300 +480+0

I = 780 kgm²

You might be interested in
Answer meeeeeeeeeeeeeee
konstantin123 [22]

Answer:

option A is correct because air friction is greater than gravity

Explanation:

hii i am boŕed wanna some fun on zòom

816 3823 4736

227UHQ

5 0
3 years ago
Read 2 more answers
The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much
aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}

Upon substituting our given values, we will get:

\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}

\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}

\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365

Taking square root of both sides, we will get:

\frac{T_1}{T_2}=\sqrt{0.0073878361246365}

\frac{T_1}{T_2}=0.0859525225030452495

\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}

\frac{T_2}{T_1}=11.634329870476699\approx 11.634

T_2=11.634\cdot T_1

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0
3 years ago
Cathode ray particles have what type of charge
zlopas [31]

Answer:

negative charge

Explanation:

3 0
3 years ago
A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu
Len [333]

Answer:

0.4 m/s²

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 1 kg

Radius (r) = 0.2 m

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Tangential acceleration (aₜ) =?

Next, we shall determine the angular acceleration (a) of the sphere. This can be obtained as follow:

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20/10

a = 2 rad/s²

Finally, we shall determine the tangential acceleration (aₜ) of the sphere. This can be obtained as follow:

The tangential acceleration (aₜ) and the angular acceleration (a) are related according to the equation:

Tangential acceleration (aₜ) = Angular acceleration (a) × Radius (r)

aₜ = ar

With the above formula, we can obtain the tangential acceleration (aₜ) as follow:

Radius (r) = 0.2 m

Angular acceleration (a) = 2 rad/s²

Tangential acceleration (aₜ) =?

aₜ = ar

aₜ = 2 × 0.2

aₜ = 0.4 m/s²

Therefore, the tangential acceleration is 0.4 m/s²

6 0
3 years ago
What medium is most effective for a school play advertisement?
DanielleElmas [232]
Flyers are best and sometimes maybe even a mass email would work better
6 0
3 years ago
Other questions:
  • Sean climbs a tower that is 71.3 m high to make a jump with a parachute. The mass of Sean plus the parachute is 81.4 kg. If U =
    13·1 answer
  • What is the primary outgoing radiation put off by the sun?
    12·1 answer
  • What happens to a wave when it moves from one medium to another?
    6·1 answer
  • A table has a height of 36.0 inches. How many centimeters is this? (Note: 1 in
    15·1 answer
  • Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a
    9·1 answer
  • A flat, circular, copper loop of radius r is at rest in a uniform magnetic field of magnitude B that extends far beyond the edge
    9·1 answer
  • A material through which charges cannot move easily?
    11·1 answer
  • At
    8·1 answer
  • Please show the work and steps. The answer is (3.3 s, 15 m/s)
    7·1 answer
  • Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!