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viktelen [127]
3 years ago
13

A 39.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to th

e edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0
Using the law of conservation of angular momentum, we have 

<span>I1 w1 = I2 w2 </span>

<span>ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 </span>

<span>ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) </span>

<span>= 1.8995 = 1.9 rad /sec ( nearly )</span>
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According to the law of universal gravitation, any two objects are attracted to each other. The strength of the gravitational fo
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C. Planet A orbits its star faster than Planet B.

Explanation:

since Planet A is closer to the star due to its gravitational force, it will orbit its star faster than planet B

hope this helps :)

5 0
3 years ago
An object moving with a speed of 5m/s comes to rest in 10s after the brakes are applied . What is the initial velocity​?
Flauer [41]

Initial velocity is 5m/s.

8 0
2 years ago
A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i
mafiozo [28]

Answer:

   t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation:  ΣF = ma

Rotation:      ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

                    ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation:     Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

         (1)                 mgSin27° - Ffr = ma  

For rotation:         (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

        (2)                Ffr = 2/3 ma

We substitute equation (2) in equation (1):

                            mgSin27° - 2/3 ma = ma

                            mgSin27° = ma + 2/3 ma

The mass gets cancelled:

                            gSin27° = 5/3 a

                            a = (3/5)(gSin27°)

                            a = (3/5)(9.8 m/s²(Sin27°))

                            a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

                           V = √2ad; where  d = 0.327m

                           V = √2(2.67 m/s²)(0.327m)

                            V = 1.32 m/s

Because V = d/t

                             t = d/V

                             t = 0.327m/1.32 m/s

                             t = 0.24 s

7 0
3 years ago
In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co
vovangra [49]

Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

Good luck!

4 0
3 years ago
Read 2 more answers
Helppppppppp plzzzzzz
seropon [69]
I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.
4 0
3 years ago
Read 2 more answers
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