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Fudgin [204]
2 years ago
11

wishing to collect weather data on a remote island, you come up with an idea for sending a weather balloon to the location. You

attach a +1 C charged object to the balloon and plan to propel the balloon to the island using an electrostatic force. Given the balloon's size and standard wind patterns en route, you realize you will need to be able to overcome an opposing air resistance of up to 100N at any point on the balloon's 750km journey to the island. How big oof a charge will you need at your location to propel the +1C balloon, even in the face if the opposing wind, up to a distance of 750 km away?
Physics
1 answer:
gtnhenbr [62]2 years ago
4 0

The magnitude of the second charge needed to propel the first charge is 6,250 C.

<h3>Magnitude of the second charge to propel the first charge</h3>

The magnitude of the second charge is calculated by applying Coulomb's law as follows;

F = kq₁q₂/r²

where;

  • k is Coulomb's constant
  • q₁ is magnitude of first charge
  • q₂ is magnitude of second charge
  • r is the distance between the charges

Fr² =  kq₁q₂

q₂ = Fr²/kq₁

q₂ = (100 x 750,000²)/(9 x 10⁹ x 1)

q₂ = 6,250 C

Thus, the magnitude of the second charge needed to propel the first charge is 6,250 C.

Learn more about charges here: brainly.com/question/18102056

#SPJ1

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ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K

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