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e-lub [12.9K]
3 years ago
14

In a perfectly inelastic one-dimensional collision between two objects, what condition alone is necessary so that the final kine

tic energy of the system is zero after the collision?
Physics
2 answers:
dem82 [27]3 years ago
8 0

Answer:

In all types of inelastic collision we know that two objects move together with same speed after collision.

Now we know that in this type of collision energy is lost due to its permanent change in shape or size

Now when two objects will collide then the total momentum of the system will remains conserved as there is no external force on the system

So here we can say that

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f

now we know that after collision if system comes to rest

then we will have

v_f = 0

m_1v_{1i} + m_2v_{2i} = 0

so here final kinetic energy is zero in all types of inelastic collision when the two objects have initial momentum zero during their motion

Crank3 years ago
5 0
<span>The condition alone that is necessary so that the final kinetic energy of the system is zero after the collision is that the objects must have momenta with the same magnitude but opposite directions.</span>
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(b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this sit
timurjin [86]

Answer:

1/2mv² = ke²

Explanation:

Let's suppose the material in question is a spring with spring constant k, mass m and position k, the kinetic energy possessed by the string will be;

K.E = 1/2mass×velocity² i.e 1/2mv²

Its elastic potential energy will be the work done on the spring when stretched which is equal to 1/2kx²

E.P = 1/2kx²

The equation describing the case where the kinetic energy is twice the elastic potential energy will be;

K.E = 2EP... 1)

Substituting the KE and EP formula into (1), we have;

1/2mv² = 2(1/2ke²)

1/2mv² = ke² which gives the required equation

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3 years ago
A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.
Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

  1. Under the assumption that the can is a closed system, the conservation law applied to the system would be: E_{in}-E_{out}=E_{change}, where E_{in} is all energy entering the system, E_{out} is the total energy leaving the system and, E_{change} is the change of energy of the system.
  2. As the purpose is to kept the beverage can at constant temperature, the change of energy (E_{change}) would be 0.
  3. The energy  that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4) where \varepsilon is the emissivity of the surface, \sigma=5.67*10^{-8}\frac{W}{m^2K} known as the Stefan–Boltzmann constant, A_S is the total area of the exposed surface, T_S is the temperature of the surface in Kelvin, T_{\infty} is the environment temperature in Kelvin.
  4. For the can the surface area would be ta sum of the top and the sides. The area of the top would be A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2, the area of the sides would be A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2. Then the total area would be A_{total}=A_{top}+A_{sides}=0.01624m^2
  5. Then the radiation heat transferred to the can would be Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W.
  6. The can would lost heat evaporating water, in this case would be Q_{out}=\frac{dm}{dt}*h_{fg}, where \frac{dm}{dt} is the rate of mass of water evaporated and, h_{fg} is the heat of vaporization of the water (2257\frac{J}{g}).
  7. Then in the conservation balance: Q_{in}-Q_{out}=Q_{change}, it would be1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0.
  8. Recall that 1W=1\frac{J}{s}, then solving for \frac{dm}{dt}:\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}
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2 years ago
Convert 25 cm to km using the method of dimensional analysis
pogonyaev

25 x 10^-5
= 0.00025

25 cm
= 0.00025 km
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