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Bad White [126]
3 years ago
8

Help with this please

Physics
1 answer:
kompoz [17]3 years ago
3 0
An airplane requires a much less force in order to get off the ground than a rocket does, since a rocket needs to exit Earth’s atmosphere.

Planes typically travel at much slower speeds that traditional rockets, as they are faced with varying restrictions that affect how quickly they fly.

Both forms of transportation, although very different in certain aspects, are similar in the fact that they both need to adhere to similar constraints. Although most airplanes aren’t faced with the challenge of exiting the atmosphere, they do need to focus on their fuel, their safety, etc. just as rockets do as well.
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A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with
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2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!
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Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements th
neonofarm [45]

Answer:

a)  q₁ = 15. 28 10⁻⁶C, b)  q₂ = 5.64 10⁻⁶ C

Explanation:

For this exercise we use Newton's second law where force is Coulomb's electric force

Case 1. Distance (x₁ = 0.200 m) from the third sphere

         F₁ = F₁₃ - F₂₃

         F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²

         F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²

Case2 Distance (x₂ = 0.6 m) from the third sphere

        F₂ = F₁₃ + F₂₃

        F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²

        F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²

The distance is between the spheres, in the annex you can see the configuration of the charge and forces

Let's replace the values

        F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²

        F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²

        6.50 = 674. 25 10³ (q₁ –q₂)

        3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)

We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second

         q₁ = q₂ + 6.50 / 674 10³

         3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04

         1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶

         q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶

         q₂ 27,777 = 156,557 10⁻⁶

         q₂ = 156.557 10-6 /27.777

         q₂ = 5.636 10⁻⁶ C

We look for q1 in the other equation

        q₁ = q₂ + 6.50 / 674 10³

        q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶

        q₁ = 15. 28 10⁻⁶C

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