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Over [174]
2 years ago
10

A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotat

ing hollow cylinder reaches a certain constant speed, v, the floor moves downward. Both passengers stay "pinned" against the wall of the rotor, as shown in the diagram below.
The magnitude of the frictional force between the adult and the wall of the spinning rotor is F. What is the magnitude of the frictional force between the child and the wall of the spinning rotor?
Physics
1 answer:
Ulleksa [173]2 years ago
8 0

Answer:

 fr ’= ½ F

Explanation:

For this exercise we use the translational equilibrium equation, on the axis parallel to the wall

              fr - W = 0

              fr = W

for the adult man they indicate that the friction force is equal to F

              F = M g

we write the equilibrium equation for the child

             fr ’= w’

             fr ’= m g

in the statement they tell us that the mass of the adult is 2 times the mass of the child

             M = 2m

we substitute

            fr ’= M / 2 g

            fr ’= ½ Mg

we substitute

            fr ’= ½ F

therefore the force of friction in the child is half of the friction in the adult

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By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l
Genrish500 [490]

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient \alpha=23.0\times10^{-6}/C^{\circ}

We need to calculate the volume expansion coefficient

Using formula of  volume expansion coefficient

\beta=3\alpha

Put the value into the formula

\beta=3\times23.0\times10^{-6}

\beta=0.000069=69\times10^{-6}/C^{\circ}

We need to calculate the volume

V= a^3

V=4^3

V=64\ cm^3

The change temperature of the cube is

\Delta T=T_{f}-T_{i}

Put the value into the formula

\Delta T=67-19 = 48^{\circ}C

We need to calculate the increases volume

Using formula of increases volume

\Delta V=V\beta\Delta T

Put the value into the formula

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5 0
3 years ago
Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.
andriy [413]

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g

where;

A = L²

\rho_{\omega} = \rho

F = ky.

Then,

Ay \rho g = ky

L^2y \rho g = ky

Divide both sides by y

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3 0
3 years ago
(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t
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Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

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Q = m * Lf

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Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

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Now, the entropy change is:

ΔS = 4,861.8 / 273

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b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

8 0
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