The answer to your question is A. Have a great day
Answer:
The horizontal component of the velocity vector is;
vh = 34.4 ft/s
The vertical component of the velocity vector is;
vy = 49.1 ft/s
Explanation:
Given;
Velocity of football v = 60 ft/s
Angle of elevation ∅ = 55°
The horizontal component of the velocity vector is;
vh = vcos∅
Substituting the values;
vh = 60cos55°
vh = 34.41458618106 ft/s
vh = 34.4 ft/s
The vertical component of the velocity vector is;
vy = vsin∅
Substituting the values;
vy = 60sin55°
vy = 49.14912265733 ft/s
vy = 49.1 ft/s
The approximate diameter of an inflated basketball is <span>2 × 10^–2 m. The answer is number 1. The rest of the choices do not answer the question above because they are too big for the diameter of the basketball.</span>
Answer:
The speed of the projectile upon returning to its starting point is 50 m/s.
Explanation:
Given;
initial velocity of the projectile, u = 50 m/s
The velocity of the projectile is maximum before hitting the ground. As the object moves upward, the velocity reduces as it approaches the maximum height, at the maximum height the velocity becomes zero. Also, as the object moves downward, the velocity starts to increase and becomes maximum before hitting the ground.
Therefore, the speed of the projectile upon returning to its starting point is 50 m/s.
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
