Answer:
v' = -2.5 cm/s
Explanation:
given,
mass of toy car, m = 10 g
speed of right, u= 20 cm/s
mass of toy car, M = 24 g
speed opposite, u' = 35 cm/s
after collision
10 g car moves to left, v = 58 cm/s
v' = ?
using conservation of momentum
m u + M u' = m v + M v'
10 x 20 - 24 x 35 = - 10 x 58 + 24 x v'
24 x v' = -60
v' = -2.5 cm/s
speed of second car after collision = -2.5 cm/s
Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
Answer:
4.28 s
Explanation:
after two seconds (2 s) His friends is
d = 3.5 m/s x 2 s = 7 meter ahead.
in this state, a bicylist start from initial velocity vo = 0 m/s and accelerat 2.4 m/s²
then, when bicylist reach his friend
t friend = t bicyclist = t
d bicylist = d friend + d
-------
d friend = 3.5 . t
d bicylist = vo . t + ½ a t²
d friend + d = vo . t + ½ a t²
3.5 t + 7 = 0 . t + ½ . 2.4 . t²
3.5 t + 7 = 1.2 t²
0 = 1.2 t² - 3.5 t - 7
t = -1.363 and t = 4.28
take the positive one