What is the formula for potassium iodide

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0

3 years ago

You are watching an object that is moving in shm. when the object is displaced 0.600 m to the right of its equilibrium position,

larisa [96]

It should take at least 4.50

8 0

3 years ago

Help! 27 degrees Celsius - What is the temperature in Fahrenheit? please help i will make you a cool thing on the computer ( gra

grandymaker [24]

The answer is 27C = 81F

8 0

3 years ago

suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round

Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

$\tau = \frac{dL}{dt}$

where L = angular momentum

since here torque is ZERO

$0 = \frac{dL}{dt}$

L = constant

so here we can write initial angular momentum of the system as

$L = (I_1 + I_2)*\omega$

here we know that

$I_1$ = moment of inertia of merry go round

$I_2$ = moment of inertia of child

so here we can say

$(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2$

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

$I_2 = mr^2$