Answer:
3. Higher in some places and lower in other places
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
Newton’s first law is commonly stated as:
An object at rest stays at rest and an object in motion stays in motion.
However, this is missing an important element related to forces. We could expand it by stating:
An object at rest stays at rest and an object in motion stays in motion at a constant speed and direction unless acted upon by an unbalanced force.
By the time Newton came along, the prevailing theory of motion—formulated by Aristotle—was nearly two thousand years old. It stated that if an object is moving, some sort of force is required to keep it moving. Unless that moving thing is being pushed or pulled, it will simply slow down or stop. Right?
This, of course, is not true. In the absence of any forces, no force is required to keep an object moving. An object (such as a ball) tossed in the earth’s atmosphere slows down because of air resistance (a force). An object’s velocity will only remain constant in the absence of any forces or if the forces that act on it cancel each other out, i.e. the net force adds up to zero. This is often referred to as equilibrium. The falling ball will reach a terminal velocity (that stays constant) once the force of air resistance equals the force of gravity.
Hope this help
Answer:
Magneta is a mix of blue and red and is a secondary colour.
Explanation:
As we can see green is the complementary colour of Magneta.Complementary colours are the pairs of colours which when combined cancel each other out. This means that when combined they produce a grayscale colour like white or black .
Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
