The Short Answer: Earth’s tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun’s most direct rays. So, when the North Pole tilts toward the Sun, it’s summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it’s winter in the Northern Hemisphere.
The velocity of the red cart after the collision is 2 m/s
From the law of conservation of momentum, initial momentum of system = final momentum of system.
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄ where m₁ = mass of red cart = 4 kg, v₁ = velocity of red cart before collision = + 4 m/s, v₃ = velocity of red cart after collision, m₂ = mass of blue cart = 1 kg, v₂ = velocity of blue cart before collision = 0 m/s (since it is initially at rest) and v₄ = velocity of blue cart after collision = + 8 m/s.
Substituting the values of the variables into the equation, we have,
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
4 kg × 4 m/s + 1 kg × 0 m/s = 4v₃ + 1 kg × 8 m/s
16 kgm/s + 0 kgm/s = 4v₃ + 8 kgm/s
16 kgm/s = 4v₃ + 8 kgm/s
16 kgm/s - 8 kgm/s = (4 kg)v₃
(4 kg)v₃ = 8 kgm/s
Divide both sides by 4 kg, we have
v₃ = 8 kgm/s ÷ 4 kg
v₃ = 2 m/s
The velocity of the red cart after the collision is 2 m/s.
Learn more about conservation of momentum here:
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Answer:
2.8351×10^7 N/kg
Explanation:
We know that,
W = mg -------(1)
where W = weight, m = mass, g = gravitational acceleration
Also from Newton's law of gravitation we know that,
F = GMm/r²
Where,
F = gravitational force,
G = universal gravitational constant
M,m = masses of object under gravitational influence
r = distance between the two center of masses.
Here we get
F = (Gm/r²)M -------(2)
From 1 and 2
The acceleration due to gravity = gravitational field intensity
So if you consider the gravitational field intensity at the surface of the asteroid,it is equal to the acceleration due to gravity at the same place.
So we get,
g = (6.67×10^-11)×9.3835×10^20/(47²) = 2.8351×10^7 N/kg
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
