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dangina [55]
2 years ago
6

A helium weather balloon is filled to a volume of 219 m3 on the ground, where the pressure is 754 torr and the temperature is 29

8 K. As the balloon rises, the pressure and temperature decrease, so it is important to know how much the gas will expand to ensure that the balloon can uphold the expansion. What is the volume at an altitude of 10,000 m, where the atmospheric pressure is 210. torr and the temperature is 230 K?
Physics
1 answer:
Novosadov [1.4K]2 years ago
8 0

Answer:

V_2 = 606.9 m^3

Explanation:

By ideal gas equation law we know that

PV = nRT

now we know that when balloon rises to certain level then the number of moles will remains same

so we can say

n_1 = n_2

\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}

now plug in all data to find the final volume of the balloon

\frac{754\times 219}{R(298)} = \frac{210 \times V_2}{R(230)}

V_2 = \frac{230\times 754 \times 219}{210 \times 298}

V_2 = 606.9 m^3

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Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

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Explanation:

a)

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  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

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