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forsale [732]
3 years ago
9

A cylinder of gas is compressed by a piston from an initial volume of 125 liters to a final volume of 90 liters. The compression

occurs at constant pressure, and the work done on the gas by the piston is 104 joules. What is the gas pressure during the compression? (1 liter = 10-3 meters3)
Physics
2 answers:
ololo11 [35]3 years ago
7 0
Since the process occurs at constant pressure, the work done by the gas on the piston is given by
W=p \Delta V = p (V_f - V_i)
where
W is the work
p is the pressure
V_f is the final volume
V_i is the initial volume

In our problem, we have
V_i =125 L = 125 \cdot 10^{-3}m^3
V_f = 90 L = 90 \cdot 10^{-3} m^3
W=-104 J (where the negative sign means the work is done by the piston on the gas)

We can re-arrange the previous equation and use these values to find the pressure:
p= \frac{W}{V_f - V_i}= \frac{-104 J}{90 \cdot 10^{-3} m^3 - 125 \cdot 10^{-3} m^3}=2971 Pa
mrs_skeptik [129]3 years ago
4 0

Answer: 3 x 10^5 pascals

Explanation: I think you meant to put 10^4 joules instead of 104, which is why nobody likes the previous answer, using 10^4 joules gets you. the answer with the following:

Using the same equation the person above me posted....

 -10^4/(90E-3-125E-3) And then only one significant figure from the question and we round to 3 x 10^5 pascals

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Question 1<br> 2 pts<br> Explain what causes a solution to be a strong acid.
lubasha [3.4K]

Answer:

Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.

5 0
3 years ago
50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in
xxTIMURxx [149]

Answer:

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m(w) = 50g

Temperature of ice, T(i) = 0° C

Temperature of water, T(w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L(w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C

m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have

50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C

3 0
3 years ago
could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad
dexar [7]

Answer:

f = 4000 / 30 sec = 133.3    vibrations/sec

P = 1 / f = .0075 sec       period of 1 vibration

4 0
2 years ago
330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°
Olenka [21]
The final temperature of the system is 32.5°
we know,  H = mcT 
where, H = Heat content of the body 
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry 

The net heat remains constant i.e. 
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x \frac{900}{4200} x (T - 10) 

⇒ 14,850 - 330T = 183.21T  - 1832 

⇒ - 513.21 T = - 16682

or T = 32.5°
3 0
3 years ago
Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha
Sedaia [141]

Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

   The correct option is b

Explanation:

From the question we are told that

          The charge carried by A is  q C

           The charge carried by B is 0 C

            The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by  A and B is mathematically evaluated as

                 \frac{ 0 + q}{2} =   \frac{q}{2}

When C and B are brought close and then separated the charge carried by  C and B  is mathematically evaluated as    

                    \frac{0 + \frac{q}{2} }{2}  = \frac{q}{4}

When C and A are brought close and then separated the charge carried by  C and A  is mathematically evaluated as  

                       \frac{\frac{q}{4} +  \frac{q}{2} }{2}   = \frac{3q}{8}

Looking at these calculation we can see that the charge carried by B is

        \frac{q}{4} C

8 0
3 years ago
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