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:
true
Explanation:
this is the answer to this question
Answer:
Explanation:
initial velocity v = 2.1 x 10⁷ m/s
acceleration a = 5.1 x 10¹⁵ m /s²
horizontal distance covered = 5.5 x 10⁻² m
time taken to cover horizontal distance = 5.5 x 10⁻² / 2.1 x 10⁷
= 2.62 x 10⁻⁹ s .
b )
vertical distance travelled due to vertical acceleration
= 1/2 a t²
= .5 x 5.1 x 10¹⁵ x (2.62 x 10⁻⁹)²
= 17.5 x 10⁻³ m
Answer:
The answer to the question is
Its maximum speed is 1.54 m/s
Explanation:
Work done = Kinetic energy
0.5·m·v² = 0.5·k·x²
Where
m = mass
v = velocity
k = spring constant
x = extension of the spring
We note that Force F is given by
F = m·a
Where
a = acceleration due to gravity
= 0.153×9.8 = 1.4994 N
Equating the work done by the force to the work done on the spring gives
Work done = Force × Distance = 1.4994×x = 0.5×k÷x² = 0.5×24.7×x²
x = 1.4994÷12.35 = 0.121 m
Substituting the value of x into the equation below gives
0.5·m·v² = 0.5·k·x²
0.5×0.153×v² = 12.35×0.121²
v² = 0.182÷0.0765 = 2.379
v = 1.54 m/s