When a loose brick is resting on a wall, it has energy. When the brick is pushed off the wall and is falling down, the amount of
2 answers:
The complete sentence is as follows:
When a loose brick is resting on a wall, it has potential energy. When the brick is pushed off the wall and is falling down, the amount of potential energy is decreasing while the amount of kinetic energy is increasing.
Explanation:
- Potential energy is the energy related to the position of an object: in particular, gravitational potential energy is given by
where m is the mass of the object, g is the gravitational acceleration, and h is the height of the object above the ground.
- Kinetic energy is the energy related to the motion of an object, and it is given by
where m is the mass and v the speed of the object.
In the example, we see that:
- at the beginning, the brick is at rest on the wall, so it only has potential energy due to its height above the ground (while the kinetic energy is zero, since it is at rest, so speed v is zero)
- as it falls down, the potential energy decreases, because the height h decreases, while the kinetic energy increases, because the speed of the object increases as it falls down.
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Answer:
The answer is "a, c and b"
Explanation:
- Its total block power is equal to the amount of potential energy and kinetic energy.
- Because the original block expansion in all situations will be the same, its potential power in all cases is the same.
- Because the block in the first case has no initial speed, the block has zero film energy.
- For both the second example, it also has the
velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher.
- While over the last case the kinetic speed is greater and lower than in the first case, the total energy is also higher than the first lower than that of the second.
- The greater the amplitude was its greater the total energy, therefore lower the second, during the first case the higher the amplitude.
Answer:
(A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
Explanation:
Given that,
Voltage = 8.2 V
Inductor = 38 mH
Resistance = 150 Ω
Time t = 0.110 ms
The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance
We need to calculate the current
Using formula of current
Put the value into the formula
(B). We need to calculate the store energy in the inductor
Using formula of energy
Put the value into the formula
{tex]E=7.04\ \mu J[/tex]
Hence, (A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
Answer:
change in internal energy 3.62*10^5 J kg^{-1}
change in enthalapy 5.07*10^5 J kg^{-1}
change in entropy 382.79 J kg^{-1} K^{-1}
Explanation:
adiabatic constant
specific heat is given as
gas constant =287 J⋅kg−1⋅K−1
specific heat at constant volume
change in internal energy
change in enthalapy
change in entropy