Its because particles of air inside it are moving faster and making more pressure plus there is a pressure from the street.
Answer: A. All of the answers are correct.
Answer:
Explanation:
Given
Ship A velocity is 40 mph and is traveling 35 west of north
Therefore in 2 hours it will travel 
thus its position vector after two hours is

similarly B travels with 20 mph and in 2 hours
![=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}](https://tex.z-dn.net/?f=%3D20%5Ctimes%202%3D40%20miles%20%3C%2Fp%3E%3Cp%3EIts%20position%20vector%5Btex%5Dr_B%3D40sin80%5Chat%7Bi%7D%2B40cos80%5Chat%7Bj%7D)
Thus distance between A and B is



Velocity of A

Velocity of B

Velocity of A w.r.t B


Angles, they line up their pool que with the pocket and make the shot
Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(
)
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [
) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ (
)
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m