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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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13 points and brainlyest if possible. Thanks.
nikdorinn [45]
Most likely it would be C not completely sure 
3 0
3 years ago
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An ideal gas occupies 600 cm3 at 20c. at what temperature will it occupy 1200 cm3 if the pressure remains constant? 10c 40c 100c
Anna11 [10]
ANS : 313℃
You need to use K in this.
To convert​ ℃ to Kelvin (K), add 273.15 to ℃.

5 0
3 years ago
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0
3 years ago
A horizontal spring with spring constant 950 N/m is attached to a wall. An athlete presses against the free end of the spring, c
OLga [1]

Answer:

66.5N

Explanation:

F = kx

Where F = force

K = spring constant

x = compression

Given

K = 950N/m

x = 7.0cm

F = ?

First convert the compression to meters .

7.0cm = 7.0 x 0.01

= 0.07 meters

Therefore

F = 950 x 0.07

= 66.5N

4 0
3 years ago
hey guys, l need some help! what is the weight of potatoes of mass 75 kg on the surface of the earth? (g=9.8m/s square) ans=735N
Airida [17]

Answer:

735N

Explanation:

Weight is a measurement of force.

F = Force

m = mass

a = acceleration

F = m*a

F = 75kg*9.8m/s²

F = 735N

8 0
3 years ago
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