Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
P = 7.28 N.s
Explanation:
given,
initial momentum of cue ball in x- direction,P₁ = 9 N.s
momentum of nine ball in x- direction, P₂ = 2 N.s
momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s
momentum of the cue after collision = ?
using conservation of momentum
in x- direction
P₁ + p = x + P₂
p is the initial momentum of the nine balls which is equal to zero.
9 + 0 = x + 2
x = 7 N.s
momentum in x-direction.
equating along y-direction
P'₁ + p = y + P'₂
0 + 0 = y + 2
y = -2 N.s
the momentum of the cue ball after collision is equal to resultant of the momentum .


P = 7.28 N.s
the momentum of the cue ball after collision is equal to P = 7.28 N.s
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Answer:
0.046
Explanation:
displacement = velocity/ time
d = 6m/s / 130s
d = 0.046m