Answer:

Explanation:
given,
mass of wheel(M) = 3 Kg
radius(r) = 35 cm
revolution (ω_i)= 800 rev/s
mass (m)= 1.1 Kg
I_{wheel} = Mr²
when mass attached at the edge
I' = Mr² + mr²
using conservation of angular momentum






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:
The minimum speed required is 5.7395km/s.
Explanation:
To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

or

where
is the mass of the asteroid,
is its distance form earth's center,
is the mass of the earth, and
is the gravitational constant.
Solving for
we get:

putting in numerical values gives


in kilometers this is

Hence, the minimum speed required is 5.7395km/s.
We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.
As it leaves from rest, it means that the initial speed is zero
t₁=9 min=540 s
t₂=10 min=600 s
x₁=at₁²/2=8*540²/2=4*291600=1166400 m
x₂=at₂²/2=8*600²/2=4*360000=1440000 m
Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel