Answer:
maximumforce is F = mg
Explanation:
For this case we must use Newton's second law,
Σ F = m a
bold indicate vectors, so we will write it in its components x and y
X axis
Fₓ = maₓ
Axis y
Fy - W = m a
Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components
Cos θ = Fₓ / F
Fₓ = F cos θ
sin θ = Fy / F
Fy = F sin θ
Let's replace and calculate
F sin θ -w = m a
As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)
F sin θ = w = mg
The maximum value of this equation occurs when the sin=1, in this case
F = mg
<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
Answer:
A) 
B) 
Explanation:
A) In this situation we are talking about a car moving only in the X- axis, hence the velocity of the car is:
Where the unit vectors 
, 
and 
represent the components 
, 
and 
in the cartesian plane.
In this sense, each unit vector is defined to have a magnitude of exactly one (1).
B) Velocity is defined as the variation of position in time, if this car is moving only along the x direction we will have:
Clearing the position:
Answer:
E. downward and constant
Explanation:
Freefall is a special case of motion with constant acceleration because the acceleration due to gravity is always constant and downward. This is true even when an object is thrown upward or has zero velocity.
For example, when a ball is thrown up in the air, the ball's velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration ggg, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.
Answer:
Explanation:
Initial velocity u = V₀ in upward direction so it will be negative
u = - V₀
Displacement s = H . It is downwards so it will be positive
Acceleration = g ( positive as it is also downwards )
Using the formula
v² = u² + 2 g s
v² = (- V₀ )² + 2 g H
= V₀² + 2 g H .
v = √ ( V₀² + 2 g H )
