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adell [148]
3 years ago
5

A 4.44 l container holds 15.4 g of oxygen at 22.55°c. what is the pressure?

Physics
1 answer:
padilas [110]3 years ago
5 0
We will use the ideal gas equation:
PV = nRT, where n is moles and equal to mass / Mr
P = mRT/MrV
P = 15.4 x 8.314 x (22.55 + 273) / 32 x 4.44
P = 266.3 kPa
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Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0
lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

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