Answer:
a) w = 9.599 10⁴ rad / s
, b) v = 3.35 10¹⁶ m / s
, c) a = 3.22 10²¹ m / s²
Explanation:
For this exercise we must use the relation of angular kinematics
a) angular velocity, the distance remembered in orbit between time (period)
w = 2π r / T
w = 2 π 3.59 10¹¹ / 2.35 10⁷
w = 9.599 10⁴ rad / s
b) linear and angular velocity are related by the equation
v = w r
v = 9,599 10⁴ 3.49 10¹¹
v = 3.35 10¹⁶ m / s
c) the centripetal acceleration is
a = v² / r = w² r
a = (9,599 10⁴)² 3.49 10¹¹
a = 3.22 10²¹ m / s²
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
= I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
Answer:
<h2>a) 50°</h2><h2>b) 40°</h2>
Explanation:
Check the complete diagram n the attachment below
a) The angle of incidence on a plane surface is the angle between the incidence ray and the normal ray acting on a plane surface. The normal ray is the ray perpendicular to the surface while the incidence ray is the ray striking a plane surface.
According to the diagram, the angle of reflection r₂ on M₂ is 90°-g where g is the angle of glance.
Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°
According the second law of reflection, the angle of incidence = angle of reflection, therefore i₂ = r₂ = 50° (on M₂)
Also ∠OO₂O₁ = ∠OO₁O₂ = 40° (angle of glance on M₁){alternate angle}
The angle of incidence on M₁ = 90° - 40° = 50°
b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°
Answer:
C. Plant A orbits its star faster than Plant B
Explanation:
Did it on study island