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kumpel [21]
3 years ago
13

A 20-kg child starts at the center of a playground merry-go-round that has a radius of 1.5 mm and rotational inertia of 500kg⋅m2

and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center.
What is its rotational speed when she gets to the edge?
Physics
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer:

w = 0.189 rad/ s

Explanation:

This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants

Initial

         L₀ = I₀ w₀

Final

          L_{f} = I w

         L₀ = L_{f}

         I₀ w₀ = I_{f} w

         .w = I₀/I_{f}   w₀

The initial moment of inertia is

        I₀ = 500 kg. m2

The final moment of inertia

         I_{f} = 500 + m r²

         I_{f} = 500 + 20 1.5

         I_{f} = 530 kg m²

Initial angular velocity

         w₀ = 0.20 rad / s

Let's calculate

          w = 500/530 0.20

          w = 0.189 rad / s

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PLEASE HELP
Sergeu [11.5K]

The vertical component of the initial velocity is v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is v_0_x = \frac{x}{t}

The horizontal displacement when the object reaches maximum height is X = \frac{xy}{gt^2} + \frac{x}{2}

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt

The horizontal component of the initial velocity is calculated as;

x = v_0_xt\\\\v_0_x = \frac{x}{t}

The time to reach to the maximum height is calculated as;

T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T =  \frac{1}{g}  (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t

The horizontal displacement when the object reaches maximum height is calculated as;

X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}

Learn more here: brainly.com/question/20689870

8 0
3 years ago
At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel
sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

7 0
3 years ago
The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
klemol [59]
The resistance is 27.5 ohms
5 0
4 years ago
Read 2 more answers
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
_____ is the total kinetic energy of particles in an object.
maw [93]
Thermal energy (or thermal kinetic energy)  is the total kinetic energy of particles in an object

hope this helps
4 0
3 years ago
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