Explanation:
A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path subjected to gravity force only.
Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.
In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.
Answer:
a. mass density
Explanation:
<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>
- When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.
<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>
<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>
Answer:
7350 J
Explanation:
The gravitational potential energy of the rock sitting on the edge of the cliff is given by:
where
m is the mass of the rock
g is the gravitational acceleration
h is the height of the cliff
In this problem, we have
m = 50 kg
g = 9.8 m/s^2
h = 15 m
Substituting numbers into the formula, we find:
Answer:
The displacement of the spring due to weight is 0.043 m
Explanation:
Given :
Mass 
Kg
Spring constant 
According to the hooke's law,
Where 
force, 
displacement
Here,
( 
)
N
Now for finding displacement,
Here minus sign only represent the direction so we take magnitude of it.
m
Therefore, the displacement of the spring due to weight is 0.043 m
