Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:
1 answer:
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Answer:
a)V = 25.1 m/s
b)V = 4.226 m/s
Explanation:
Given that
x(t)=A t + B t²
A = -4.3 m/s
B = 4.9 m/s²
x(t)= - 4.3 t +4.9 t²
The velocity of the particle is given as
V=-4.3 + 4.9 x 2 t
V= - 4.3 + 9.8 t m/s
Velocity at point t= 3 s
V= - 4. 3 + 9.8 x 3 m/s
V= - 4.3 + 29 .4 m/s
V = 25.1 m/s
At origin :
x= 0 m
0 = - 4.3 t +4.9 t²
0 = - 4.3 + 4.9 t
t=0.87 s
The velocity at t= 0.87 s
V= - 4.3 + 9.8 t m/s
V= - 4. 3 + 9.8 x 0.87 m/s
V= - 4.3 + 8.526 m/s
V = 4.226 m/s
a)V = 25.1 m/s
b)V = 4.226 m/s
Answer:
a) I = ( +
) L² , b) w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
Explanation:
a) The moment of inertia is a scalar that represents the inertia in circular motion, therefore it is an additive quantity.
The moment of inertia of a rod held at one end is
I₁ = 1/3 M L²
The moment of inertia of the mass at y = L
I₂ = m y²
The total inertia method
I = I₁ + I₂
I = \frac{1}{3} M L² + m (\frac{2}{3} L)²
I = ( +
) L²
b) The conservation of angular momentum, where the system is formed by the masses and the bar, in such a way that all the forces during the collision are internal.
Initial instant. Before the crash
L₀ = I₂ w₀
angular and linear velocity are related
w₀ = y v₀
w₀ = L v₀
L₀ = I₂ y v₀
Final moment. After the crash
= I w
how angular momentum is conserved
L₀ = L_{f}
I₂ y v₀ = I w
substitute
m ()² (\frac{2L}{3} v₀ = (
+
) L² w
m L³ v₀ = (
+
) L² w
m L v₀ = (
+
) w
L v₀ = w
w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀