One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion.
BTW, circles are ellipses too, but so special that their likelihood is close to zero.
Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4
Explanation:
devide 20/6 and 20/5 respectively.
Answer:
Explanation:
kinetic energy required = 1.80 MeV
= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 2.88 x 10⁻¹³ J
If v be the velocity of proton
1/2 x mass of proton x v² = 2.88 x 10⁻¹³
= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³
v² = 3.45 x 10¹⁴
v = 1.86 x 10⁷ m /s
If V be the potential difference required
V x e = kinetic energy . where e is charge on proton .
V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³
V = 1.8 x 10⁶ volt .
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
