The statement “the scientific process is open-ended “means
2 answers:
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Answer:
a) v² = G M R³, b) T = 2π /, c)
Explanation:
a) The kinetic energy is
K = ½ m v²
to find the velocity let's use Newton's second law
F = m a
acceleration is centripetal
a = v² / R
force is the universal force of attraction
F = G m M / r²
we substitute
G m M R² = m v² R
v² = G M R³
the kinetic energy is
K = ½ m G M R³
b) angular and linear velocity are related
v = w R
w = v / R
w =
w =
the angular velocity is related to the period
w = 2π / T
T = 2π / w
we substitute
T = 2π /
c) the angular moeomto is
L = m v r
L = m RA G M R³ R
L =
Answer:
The coefficient of friction is (F/(19.6·m)
Explanation:
The given parameters are;
The force applied to the immovable body = F
The time duration the force acts = t
The time the body spends in motion = 3·t
The acceleration due to gravity, g = 9.8 m/s²
From Newton's second law of motion, we have;
The impulse of the force = F × t = m × Δv₁
Where;
Δv₁ = v₁ - 0 = v₁
The impulse applied by the force of friction, is
× (3·t - t) =
× (2·t)
Given that the motion of the object is stopped by the frictional force, we have;
The impulse due to the frictional force = Momentum change = m × Δv₂ = × (2·t)
Where;
Δv₂ = v₂ - 0 = v₂
Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to act, v₁, we have;
m × Δv₂ = × (2·t) = m × Δv₁ = F × t
∴ × (2·t) = F × t
= F × t/(2·t) = F/2
The coefficient of dynamic friction, = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)
= (F/(19.6·m)
The coefficient of friction, = (F/(19.6·m)