Answer:
15.66 rad/s
Explanation:
The vertical motion and horizontal motion are independent of each other.
t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m
t = √ ( 36 / 9.81 ) = 1.916 secs
horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)
tangential velocity V = angular velocity ( ω) × radius
distance traveled = ω × r × t = 30 × r
radius cancelled on both side
ω = 30 / 1.9156 = 15.66 rad/s
The gravitational force between Mars and the Sun is 
Explanation:
The magnitude of the gravitational force between two objects is given by the equation:
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, we have:
is the mass of the Sun
is the mass of Mars
is the average distance Mars-Sun
Substituting into the equation, we find the gravitational force:

So, the closest answer is

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Answer:
False
Explanation:
When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.
As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.
ωd = ω₀√(1 - ζ)
Where ζ is called damping ratio.
For small value of ζ
ωd ≈ ω₀
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2
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