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Answer:
He made great advancements in developing a logical way to know more about the universe and celestial entities inside the space. And this theory is termed to be heliocentric in nature.
Explanation:
- In early times most of the people believed that our planet Earth is the center of the universe or the solar system and rest of the celestial entities move around it in a given path, so, it confused the well known scientist named as Galileo Galilei. As, he observed the different dark patches or shadow like textures on the face of the Sun.
- While, it is more obvious to known that any object having multiple small shadows means that it is present inside such a region that all of the celestial entities move or orbit around it in a given way.So, he concluded that planet Earth itself move around the red Giant in a given way rather then being the center of the universe.
Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
