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Alexxx [7]
3 years ago
13

A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have

masses of 15 kg, 18 kg, 22 kg, and 25 kg. They start out on the edge of the merrygo-round but later move toward the center until they are halfway between the edge and the center.
A. Sketch the arrangement of the children on the merry-go-round when they are at the edge and halfway to the middle.
B. What is the total moment of inertia for both arrangements?
C. The children are standing at the edge when their parents get the ride turning at 0.21 radians/sec. What is their tangential velocity and the period of rotation?
D. While the ride rotates the children move to their positions halfway to the center. What is the angular velocity when they get to their new spots? Explain.

Physics
1 answer:
mixer [17]3 years ago
5 0

Answer:

B) I1 = 1680 kg.m^2          I2 = 1120 kg.m^2

C) V = 0.84m/s      T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2   where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2    Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

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7 0
2 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
A horizontal compass is placed 21 cm due south from a straight vertical wire carrying a 36 a current downward. in what direction
Anit [1.1K]

 <span>
The needle of a compass will always lies along the magnetic field lines of the earth. 
A magnetic declination at a point on the earth’s surface equal to zero implies that 
the horizontal component of the earth’s magnetic field line at that specific point lies along 
the line of the north-south magnetic poles. </span>

The presence of a current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field. Since magnetic 
<span>fields are vector quantities, therefore the magnetic field of the earth and the magnetic field of the vertical wire must be combined vectorially. </span></span>

<span>
Where:</span>

B1 = magnetic field of the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T

B2 = magnetic field due to the straight vertical wire along the y-axis

We can calculate for B2 using Amperes Law:

B2 = μ₀ i / [ 2 π R ]

B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>

The angle can be calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>

θ = 53°

<span>
<span>The compass needle points along the direction of 53° west of north.</span></span>

8 0
3 years ago
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