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Nastasia [14]
1 year ago
12

if 162.4 j of heat are added to a gas contained within a cylinder and the gas expands, doing 87.3 j of work, what is the change

in internal energy of the system?
Physics
1 answer:
labwork [276]1 year ago
6 0

The changes in internal energy of the system <u>75.1 J.</u>

The internal energy of a thermodynamic system is the total energy contained in it. It is the energy required to create or prepare a system in a given internal state and includes contributions of potential energy and internal kinetic energy.

Calculation:

Internal energy =  162.4 j - work done

                        =  162.4 j - 87.3 j

                         =<u> 75.1 J</u>

<u />

Internal energy, in thermodynamics, is the property or state function that defines the energy of matter in the absence of capillary action or external electric, magnetic, or other fields.

Internal energy is the microscopic energy contained in the matter given by the random and disordered kinetic energy of the molecules. It also includes the potential energy between these molecules and the nuclear energy contained in the atoms of these molecules.

Learn more about internal energy here:-brainly.com/question/15735187

#SPJ4

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Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first d
Alexxx [7]

Answer:

0.91437 m

0.22859 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s=81\ inches=81\times 0.0254=2.0574\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.0574=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.0574\times 2}{9.81}}\\\Rightarrow t=0.64764\ s

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.64764}{3}\\\Rightarrow t'=0.21588\ s

For second drop time is given by

t''=2t'\\\Rightarrow t''=2\times 0.21588\\\Rightarrow t''=0.43176\ s

Distance from second drop

s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.43176^2\\\Rightarrow s=0.91437\ m

Distance from second drop is 0.91437 m

Distance from third drop

s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.21588^2\\\Rightarrow s=0.22859\ m

Distance from third drop is 0.22859 m

6 0
3 years ago
2. A uniform wire of resistance R is stretched until its length doubles. Assuming its density and resistivity remain constant, w
never [62]

Answer:

Resistance is quadrupled.

Explanation:

Solving this requires us to use the formula of resistivity.

Resistivity is usually said to be the measure of the resistance of a particular size of any given material to the electrical conduction. It is mathematically represented as

ρ = RA/L, where

ρ = the resistivity of the given material

R = the resistance of the material

A = the area of the material

L = length of the material.

From the question, we're told that the length is doubled with the resistivity and density remaining constant. If the density is constant, this makes the volume constant as well.

Volume, V = A * L. We're then told that the length is doubled. If the length is doubled, for the volume to remain constant, then the area must be halved.

Volume, V = A/2 * 2L

Making, Resistance R, subject of the formula, we have

R = ρL/A.

Since resistivity is constant and the area is halved, we then have

R = 2L / (1/2A)

R = 4L / A

If the length is doubled, we have the resistance to be quadrupled

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3 years ago
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A 60-kg person is traveling in a car moving at 16 m/s, when the car hits a barrier. The person is not wearing a seat belt, but i
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Using conservation of momentum, we can solve for the force that the air bag exerts on the person.

Recall the equation for momentum (p):

p = mv = F*dt

We can solve for total momentum, then divide by out time interval. This gets us:

(60kg)(16m/s) = F(0.2s)&#10;&#10;&#10;F = 4800N

F = 4800N


4 0
3 years ago
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