A straight wire carrying a 3.30-A current is placed in a uniform magnetic field of magnitude 0.272 T directed perpendicular to t
1 answer:
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Answer:
Explanation:
This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.
When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.
<h3>Elastic potential energy</h3>We know that a spring following Hooke's Law has a elastic potential energy:
where is the displacement from the relaxed length and k is the spring's constant.
To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :
as we need 9.12 N to compress 4.60 cm, this means:
So, the elastic energy of the compressed spring is:
And when the spring is relaxed, the elastic potential energy will be zero.
<h3>Gravitational potential energy</h3>To see how much gravitational potential energy will the pellet win, we can use
where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.
Taking all this together, the gravitational potential energy when the spring is relaxed will be:
We know that the kinetic energy for a mass m moving at speed v is:
so, for the pellet will be
By conservation of energy, we know:
So
Answer:
The heat capacity for the second process is 15 J/K.
Explanation:
Given that,
Work = 100 J
Change temperature = 5 k
For adiabatic process,
The heat energy always same.
We need to calculate the number of moles and specific heat
Using formula of heat
Put the value into the formula
We need to calculate the heat
Using formula of heat
Put the value into the formula
We need to calculate the heat capacity for the second process
Using formula of heat
Put the value into the formula
Hence, The heat capacity for the second process is 15 J/K.