Answer:
d_2 = 4d_1
Explanation:
The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by
R = U²sin2θ/g
Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be
d_1 = U²sin2θ/g
Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be
d_2 = V²sin2θ/g
= (2U)²sin2θ/g
= 4U²sin2θ/g
= 4d_1 (since d_1 = U²sin2θ/g)
So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.
Answer:
When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.
