8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At
1 answer:
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Answer:
(a) A = 1 mm
(b)
(c)
(c) The formula for the maximum acceleration is given by
[tex]a_{max}=606.4 m/s^{2}/tex]
Answer:
G = 6,786 10⁻¹¹ m³ / s² kg
Explanation:
The law of universal gravitation is
F = G m M/ r²
Where G is the gravitational constant, m and M are the masses of the bodies and r is the distance from their centers
Let's use Newton's second law
F = m a
The acceleration is centripetal
a =
We replace
G m M / r² = m
G = r² / M
Let's replace and calculate
G = 2.7 10⁻³ (3.88 10⁸)² / 5.99 10²⁴
G = 6,786 10⁻¹¹ m³ / s² kg
Let's perform a dimensional analysis
[N m²/kg²] = [kg m/s² m² / kg²] = [m³ / s² kg]
Answer:
w = 0.189 rad/ s
Explanation:
This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants
Initial
L₀ = I₀ w₀
Final
= I w
L₀ = L_{f}
I₀ w₀ = I_{f} w
.w = I₀/I_{f} w₀
The initial moment of inertia is
I₀ = 500 kg. m2
The final moment of inertia
= 500 + m r²
I_{f} = 500 + 20 1.5
I_{f} = 530 kg m²
Initial angular velocity
w₀ = 0.20 rad / s
Let's calculate
w = 500/530 0.20
w = 0.189 rad / s