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
dem82 [27]
3 years ago
13

A ballet student with her arms and a leg extended spins with an initial rotational speed of 0.90rev/s . As she draws her arms an

d leg in toward her body, her rotational inertia becomes 0.80 kg⋅m2 and her rotational velocity is 4.1rev/s .
Determine her initial rotational inertia.
Physics
1 answer:
Rainbow [258]3 years ago
5 0
Initial inertia = 3.6 kg . m^2

final inertia = 0.80 kg.m^2rotational velocity final = 4.1 rev/s
rotational velocity initial = 0.90 rev/s 
lo= lf(final inertia) *  Wf(rotational velocity final)/ Wo (rotational velocity initial)

lo =  0.8*4.1 / 0.90 = 3.6 kg m2
You might be interested in
Young's Modulus refers to changes in the: a- Volume b- Length c- Body layers
bekas [8.4K]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

Stress is defined as force per unit area and strain is defined as proportional deformation in a material.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight

l = length of wire

A = area of cross section

\Delta l = change in length

Hence, the correct answer is Option b.

6 0
4 years ago
Muscles make up ore than __ of a person body?<br>a.10<br>b.25<br>c40<br>d.75
exis [7]
40 percent of your body
8 0
3 years ago
Read 2 more answers
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

5 0
3 years ago
The frequency of a wave is 560 Hz. What is it’s period
Crazy boy [7]
The answer would be, "1/560 seconds".
4 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
Other questions:
  • An object is attached to a vertical spring and bobs up and down between points a and
    13·1 answer
  • True or False: If a point charge has electric field lines that point into it, the charge must be positive.
    5·1 answer
  • Which of the following is the best name for CaF2?
    5·2 answers
  • What two simple machines make up an axe?
    11·1 answer
  • If we approximate the rack to be completely flat and the racecar is travelling a constant 30.5 m/s around the turn, what forces
    14·1 answer
  • In science work is defined as
    7·1 answer
  • What is the characteristic that allows an object to have kinetic energy?
    13·1 answer
  • An 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching th
    7·1 answer
  • A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s
    14·1 answer
  • An unstoppable object is heading right toward an unmovable object. What's going to happen?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!