Two carts (m1 = m2 = 0.400 kg) are placed on an aluminum track. The first cart is pushed with the initial velocity of 1.5 m/s to
wards the second cart, which remains at rest until the collision. Determine the velocities of both carts after the collision. (5)
V1 = __________ m/s, V2 = __________ m/s,
V1 = __________ m/s, V2 = __________ m/s,
1 answer:
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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:
Explanation:
Given that,
Initial velocity of an object, u = 22 m/s
Final velocity of an object, v = 36 m/s
Time, t = 5 s
It can be assumed to find the average acceleration of the object instead of average velocity.
The change in velocity per unit time is equal to average acceleration of an object. It can be given by :
So, the acceleration of the object is .