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nataly862011 [7]
3 years ago
5

A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this

process, it absorbs 500 J of heat from the surroundings.
Required:
a. Calculate the energy change of the gas, DU.
b. Calculate the work, w, done on the gas in an irreversible adiabatic (q =0) process connecting the same initial and final.
Physics
1 answer:
mars1129 [50]3 years ago
7 0

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

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hammer [34]
Work= force*distance
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Force= mass*acceleration
Force= 5 kg*6
Force= 40 N
Work= 40×12
Work= 480 J (joules)
I think this is it
5 0
3 years ago
A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
astraxan [27]

Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

Explanation:

Given:

mass of car, m=1380\ kg

A)

frequency of spring oscillation, f=1.6\ Hz

We knkow the formula for spring oscillation frequency:

\omega=2\pi.f

\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f

\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6

k_{eq}=139469.3537\ N.m^{-1}

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

k=\frac{k_{eq}}{4}

k=\frac{139469.3537}{4}

k=34867.3384\ N.m^{-1}

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }

\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }

\omega'\approx84\ Hz

7 0
3 years ago
What is the direction of the force for a negative charge moving downward in a magnetic field pointing to the left?
velikii [3]

Answer: A

Out of the screen

Explanation:

Using right hand rule, the magnetic force is perpendicular to the plane form by the magnetic field of a charged particle and its speed. Which will be into the screen.

But the negative charged particle moves in the opposite direction of the positive charged particle. Therefore, the magnetic force direction will be out of the screen

5 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorous and oxygen act together as on
S_A_V [24]

The question is incomplete, the complete question is;

The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorus and oxygen act together as one charged particle, which is connected to magnesium, the other charged particle. What does the 2 mean in the formula 5Mg3(PO4)2? A. There are two elements in magnesium phosphate. B. There are two molecules of magnesium phosphate. C. There are two magnesium ions in a molecule of magnesium phosphate. D. There are two phosphate ions in a molecule of magnesium phosphate.

Answer:

There are two phosphate ions in a molecule of magnesium phosphate.

Explanation:

The compound magnesium phosphate is an ionic compound. Ionic compounds always consists of two ions, a positive ion and a negative ion.

In this case, the positive ion is Mg^2+ while the negative ion is PO4^3-.

The subscript, 2 after the formula of the phosphate ion means that there are two phosphate ions in each formula unit of magnesium phosphate.

5 0
2 years ago
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