<span>If a plane has a velocity of 300 km/h and a tailwind of 20 km/h, then the vectors of both forces would add (assuming that the tailwind is blowing exactly at the airplanes back) to a total of 320 km/h. Hope it helps
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Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4

To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;

Substituting the values into the formula, we have;

<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
The magnitude of charge on a proton and electron is the same, 1.602 x 10-19 C. Protons are +, and electrons -.
Answer:
We know from the basic speed distance relation that

Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car

Which clearly exceeds the limit of 