It is overhead at the equator, it is because the sun ray’s
will be moving vertically as this will be directed at the equator. It is
because if it moves vertically, it will hit or overhead the equator and this
usually happens in spring and fall.
These objects would be classified as extreme trans Neptunian object (ETNO).
Explanation:
ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.
Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)
The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.
The magnitude of the force exerted by the left cube on the right cube is 17.64N.
<h3>What is frictional force?</h3>
When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.
Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.
From the equilibrium of forces in vertical direction
Normal force N= 2m x g
friction force f = μN =μ(2m)g
From the equilibrium of forces in horizontal direction
F₁ =ma =0
using Newton's third law of motion, we get
F₁ - f =0
F₁ =f = μ(2m)g
Put the values, we get
F₁ = 17.64N
Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.
Learn more about friction force.
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To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
