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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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patriot [66]

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Longer pendulums swing slower.

You need to <em>make your pendulum slightly longer</em>.

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6 0
3 years ago
Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its prev
masha68 [24]

Answer:

The speed in the smaller section is 43.2\,\frac{m}{s}

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

\Delta Q=0 (1)

With Q the flux of water that is Av with A the cross section area and v the velocity, so by (1):

A_{2}v_{2}-A_{1}v_{1}=0

subscript 2 is for the smaller section and 1 for the larger section, solving for v_{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}} (2)

The cross section areas of the pipe are:

A_{1}=\frac{\pi}{4}d_{1}^{2}

A_{2}=\frac{\pi}{4}d_{2}^{2}

but the problem states that the diameter decreases 86% so d_{2}=0.86d_{1}, using this on (2):

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1}

v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s}

5 0
3 years ago
PLEASE HELP ME PLEASE
iVinArrow [24]

Answer:

b should be the answer

Explanation:

its talking about the weather and not about the climate

7 0
2 years ago
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Answer:

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Explanation:

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3 years ago
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Arturiano [62]

Answer:

It depends on the environment you're in but I'd go with warm

Explanation:

4 0
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