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saveliy_v [14]
3 years ago
8

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s . The wheel can be considered

a uniform disk of mass 5.0 kg and diameter 0.36 m . The potter then throws a 3.1-kg chunk of clay, approximately shaped as a flat disk of radius 8.0 cm , onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it? Ignore friction.
Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

ωf =1.78 rev/s

Explanation:

it is a non elastic collision, so conservation of angular momentum is used.  and there id no friction

I₁ωi₁ + I₂ωi₂ = (I₁ + I₂)ωf

ωi₁ = 2.0 rev/s*(2π) rad/rev = 4.0π rad/s

assume ωi₂ = 0

Sol the equation to wf

ωf =  (½m₁r₁²)ωi₁ /  (½m₁r₁² + ½m₂r₂²)

ωf = (m₁r₁²)*ωi₁  /  (m₁r₁² + m₂r₂²)

ωf = (5.0kg*(0.18m)²)4.0 π rad/s / ((5.0kg*(0.18m)²) + (3.1*(0.08m)²))

ωf =11.19 rad/s/ 2π

ωf =1.78 rev/s

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Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight  due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ =  400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

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Identify the factors that cause melting and thus the formation of magma. Choose one or more: A. decompression B. compression C.
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Answer: C Heat Transfer, E Addition of Volatiles and A Decompression

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If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?
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Answer:

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2 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
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Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

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