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Temka [501]
3 years ago
11

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

at a pressure of 2.65 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm.
what is the final pressure of the system in atm?
Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1+P_2V_2=P_fV_f

where,

P_1 = first pressure = 8.19 atm

P_2 = second pressure = 2.65 atm

V_1 = first volume = 2.14 L

V_2 = second volume = 9.84 L

P_f = final pressure = ?

V_f = final volume = 2.14 L  + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L

P_f=3.64atm

Therefore, the final pressure of the system in atm is, 3.64 atm

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STatiana [176]

Answer:

1. 31,536,000 seconds

2. Car B traveled a longer distance

3. Volume of box = 0.235887 cubic meters

Explanation:

Q1. Age in seconds

1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

Therefore 1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

In scientific notation this would be 3.1536\times  10^7 \text{ seconds}

Q2. Comparing km and miles

Given:
1 km = 1000m and 1 m = 3.3ft,   I km = 1000 x 3.3 = 3300 ft.

Convert Car A distance of 25.7km to feet :
25.7 km. = 25. 7 x 3300 ft. = 84,810 ft.

For Car B  that traveled 20 miles,
20 miles = 20 x 5280 = 105,600 ft.

Since 105,600 > 84,810, car B traveled a longer distance

Q3. Volume of wooden box

The wooden box is in the shape of a rectangular prism
It volume is L x W x H
Volume = 1.525 x 0.30 x 0.5156 = 0.235887 cubic meters

5 0
10 months ago
2. A car that starts from rest can travel a distance of 50 m in a time of 6.0 s.
Helen [10]
Answer is B hope this helps
8 0
1 year ago
If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch the same
Angelina_Jolie [31]

Answer:W=\frac{3}{4} ft-lb

Explanation:

Given

Work required to stretch 1 ft is 12 ft-lb

and we have to find work required to stretch 3 in.

i.e. \frac{1}{4} ft

12=\frac{1}{2}K\left ( 1\right )^2 ------(1)

W=\frac{1}{2}k\left ( \frac{1}{4}\right )^2-----(2)

divide (1)&(2)

\frac{12}{W}=\left ( \frac{4}{1}\right )^2

W=\frac{12}{4\times 4}

W=\frac{3}{4} ft-lb

6 0
3 years ago
Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t
Marysya12 [62]

Answer:

<h2>1/7 kg</h2>

Explanation:

Find the diagram attached for better understanding of the question.

Given the mass of one of the blocks to be 1.0kg and accelerates downward at 3/4g.

g = acceleration due to gravity.

Let the block accelerating downward be M, mass of the other body be 'm' and the acceleration of the body M be 'a'.

M = 1.0 kg and a = 3.4g

According to newton's second law; \sum fy = ma_y

For body of mass m;

T - mg = ma ... (1)

For body of mass M;

Mg - T = Ma ... (2)

Adding equation 1 ad 2;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into the resulting equation;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

28m = 4

m = 1/7 kg

Therefore the mass of the other box is 1/7 kg

3 0
3 years ago
George walks to a friends house. He walks 750 meters north, then realizes he walked too far. He turns around and walks 250 meter
Nostrana [21]

Answer: 77 m/s in 2 significant figures

Explanation:

Total distance traveled (d) = 750 + 250 = 1000m

Time taken (t) = 13 s

Speed (s) = ?

Now,

s = d/t

s = 1000/13

s = 77 m/s in 2 significant figures

4 0
3 years ago
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