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omeli [17]
1 year ago
12

A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 935

°C. The other side is evacuated and has a volume twice the size of the part containing the gas. The partition is now removed and the gas expands to fill the entire tank. Heat is now transferred to the gas until the pressure equals the initial pressure. Determine the final temperature of the gas.
Physics
1 answer:
a_sh-v [17]1 year ago
7 0

Answer:

2805 °C

Explanation:

If the gas in the tank behaves as ideal gas at the start and end of the process. We can use the following equation:

P=RTn/V

The key issue is identify the quantities (P,T, V, n) in the initial and final state, particularly the quantities that change.

In the initial situation the gas have an initial volume V_{i}, temperature T_{i}, and pressure P,.

And in the final situation the gas have different volume V_{f} and temeperature T_{f}, the same pressure P,, and the same number of moles n,.

We can write the gas ideal equation for each state:

P=RT_{i}n/V_{i} and P=RT_{f}n/V_{f}, as the pressure are equals in both states we can write

RT_{i}n/V_{i} = RT_{f}n/V_{f}

solving for T_{f}

T_{f} = T_{i}/V_{i} * V_{f} (*)

We know T_{i}  = 935 °C, and that the V_{f} (the complete volume of the tank) is the initial volume V_{i} plus the part initially without gas which has a volume twice the size of the initial volume (read in the statement: the other side has a volume twice the size of the part containing the gas). So the final volume  V_{f}= V_{i} + 2V_{i}=3V_{i}

Replacing in (*)

T_{f} = 935/V_{i} * 3V_{i} = 935*3= 2805

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a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis
KatRina [158]
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m +  201.67 m = 465.67 or  approximately 4.7×10² m.
7 0
1 year ago
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2
docker41 [41]

1) At x = 6.6 cm,  E_x=3.47\cdot 10^6 N/C

2) At x = 6.6 cm, E_y=0

3) At x = 1.45 cm, E_x=-3.76\cdot 10^6N/C

4) At x = 1.45 cm, E_y=0

5) Surface charge density at b = 4 cm: +62.75 \mu C/m^2

6) At x = 3.34 cm, the x-component of the electric field is zero

7) Surface charge density at a = 2.9 cm: +65.25 \mu C/m^2

8) None of these regions

Explanation:

1)

The electric field of an infinite sheet of charge is perpendicular to the sheet:

E=\frac{\sigma}{2\epsilon_0}

where

\sigma is the surface charge density

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.

So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:

E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}

where

\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2

The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).

So,

E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C

And the negative sign indicates that the direction is to the right.

2)

We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).

So the total field along the y-direction is zero.

This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), dE_y, is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis, -dE_y. Therefore, the net field along the y-direction is always zero.

3)

Here it is similar to part 1), but this time the point is located at

x = 1.45 cm

so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is

E=E_1-E_2

Substituting the same expressions of part 1), we find

E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C

where the negative sign indicates that the direction is to the left.

4)

This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.

5)

Here we note that the slab is conductive: this means that the charges in the slab are free to move.

We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).

The net charge per unit area of the slab is

\sigma=+64\mu C/m^2

And this the average of the surface charge density on both sides of the slab, a and b:

\sigma=\frac{\sigma_a+\sigma_b}{2} (1)

Also, the infinite sheet located at x = 0, which has a negative charge \sigma_1=-2.5\mu C/m^2, induces an opposite net charge on the left surface of the slab, so

\sigma_a-\sigma_b = +2.5 \mu C/m^2 (2)

Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:

\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2

6)

Here we want to calculate the value of the x-component of the electric field at

x = 3.34 cm

We notice that this point is located inside the slab, because its edges are at

a = 2.9 cm

b = 4.0 cm

But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero

7)

We  calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is \sigma_a = +65.25 \mu C/m^2

8)

As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is

2.9 cm < x < 4 cm

So the correct answer is

"none of these region"

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0
2 years ago
Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to
Dmitriy789 [7]

Answer:

Explanation: idk sry

7 0
2 years ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

      also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360°  = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

5 0
2 years ago
A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
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