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lesantik [10]
4 years ago
5

A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe

r as an integer. Part B
How much work is done by external forces?
Express your answer as an integer.
Physics
1 answer:
Elina [12.6K]4 years ago
5 0

A) +5 J

B) +1 J

Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

\Delta U=-5 J is the loss in potential energy of the system

\Delta K=+6 J is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

E=K+U

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

\Delta E= \Delta K + \Delta U

In this problem:

\Delta K=+6 J

\Delta U=-5 J

So, the change in mechanical energy is:

\Delta E=+6+(-5)=+1 J

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

W=\Delta E=+1 J

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Explanation:

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I_o = Reference sound intensity

According to question, sound-reflecting windows that reduce the sound intensity level by 30.0 db .

Let the intensity of the sound be I_o and reduced intensity be I.

I=\frac{I_o}{30} dB

L_I=10\log\frac{I}{I_o} dB

L_I=10\log\frac{\frac{I_o}{30}}{I_o}=10\log\frac{1}{30} dB=-14.77 dB

The sound intensity was reduced by -14.77 dB.

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Why is an iron core needed in an electromagnet?
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The magnetic dipole moment of Earth is 8.00 ✕ 1022 J/T. Assume that this is produced by charges flowing in Earth's molten outer
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It is usually defined by,

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Where,

\mu = Magnetic Dipole Moment

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In this way re-arrange for the current and replacing,

I = \frac{\mu}{A}

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3 years ago
As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pres
Kaylis [27]

Answer:

Vi = 0.055 m³ = 55 L

Explanation:

From first Law of Thermodynamics, we know that:

ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

Vi = 0.058 m³ - 0.003 m³

<u>Vi = 0.055 m³ = 55 L</u>

7 0
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