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Vladimir79 [104]
4 years ago
11

A vertical spring with stiffness k originally is at rest with no mass attached. Then, a mass M is attached, and the spring rocks

back and forth until it is at rest at a new equilibrium position. What is this position in terms of g, M, and k?
Physics
1 answer:
Alina [70]4 years ago
8 0

Answer:

The position of the spring in terms of g, m & k is x = \frac{m g}{k}

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ x = \frac{m g}{k}

This is the position of the spring in terms of g, m & k.

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A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e
gayaneshka [121]

Answer:

  V_inside = 36 V

Explanation:

<u>Given  </u>

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.  

<u>Required</u>

We are asked to calculate the potential at the centre of the sphere  

<u>Solution</u>

The potential energy due to the sphere is given by equation

V = (1/4*π*∈o) × (q/r)                                          (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V  

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside                               (2)

Now we can plug our values for r, R and V_outside into equation (2) to get  V_inside

V_inside = (1.2 m )/(0.600)*18

               = 36 V

  V_inside = 36 V

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3 years ago
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How much heat is released when 12.0 grams of helium gas condense at 2.17 K? The latent heat of vaporization of helium is 21 J/g.
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To determine the heat released by the process of condensation, we simply multiply the amount of the gas that condensed to the latent heat of vaporization. We do as follows:

Heat released = 21 J/g (12.0 g ) = 252 J of heat released
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3 years ago
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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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