Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
As we know that when charge is released in electric field
It will have two forces on it
1. electrostatic force
2. gravitational force
now if the ball will accelerate upwards so we can say
net upward force = mass * acceleration
now we can find charge q on it by above equation
So above is the charge on the particle
In my estimation I would say C, I was leaning towards A, but I believe that would merely be "incomplete combustion." I hope this was semi-helpful!
Answer:
θ = (7π / 3) rad
Explanation:
given,
displacement of simple harmonic motion along x-axis
equation is given as
x = 5 sin (π t + π/3 )
general equation of simple harmonic motion
x = A sin θ
θ is the phase angle
θ = π t + π/3
at t = 2 s
Phase of the motion at t =2 s is θ = (7π / 3) rad
<span>Velocity is a vector and it has both speed and direction. It takes a force to change direction just as it does to change speed</span>. In order to have a constant velocity the object must maintain a constant direction and speed. Hope this answers the questions. Have a nice day.
