The circuit is no longer closed.
Answer:
if I aint wrong it would 2nd one
Answer:
U = 8.30×10-⁹J
Explanation:
m1 = m2 = 5.00kg masses of the spheres
d = 15.0cm = 15×10-²m
r = 5.10cm = 5.10×10-²m
R = d + r = 15×10-² + 5.10×10-²
R = 20.10 ×10-²m = 0.201m
G = 6.67×10-¹¹Nm²/kg²
U = Gm1×m2/R = potential energybetween the spheres
U = 6.67×10-¹¹×5.00×5.00/0.201
U = 8.30×10-⁹J
Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?
The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m
Answer:
Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.
