Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.
PE = 81 * 9.8 * 3.8 = 3016.44 J
Work = 1/2 * 1888 * d^2
PE = Kinetic energy at the base.
1/2 * 1888 * d^2 = 3016.44
d = 1.78 approx 1.8
F = Ke = 1888 * 1.8 = 3398.4N
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.
Is there any chemical names listed ?
D = distance between th two trains at the start of the motion = 100 miles
V = speed of the faster train towards slower train = 60 mph
v = speed of the slower train towards faster train = 40 mph
t = time taken by the two trains to collide = ?
time taken by the two trains to collide is given as
t = D/(V + v)
t = 100/(60 + 40) = 1 h
v' = speed of the bird = 90 mph
d = distance traveled by the bird
distance traveled by the bird is given as
d = v' t
d = 90 x 1
d = 90 miles
