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
1. 12.75 J
Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is
where
F = 15 N is the magnitude of the force
d = 85 cm = 0.85 m is the displacement of the cart
Substituting in the formula, we get
2. 10.6 N
In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).
Therefore, the work done is
However, in this case the displacement is
d = 120 cm = 1.20 m
Therefore, the magnitude of the force in this case is
Answer:
20 m/s
Explanation:
Recall that one of the equations of motions can be written:
v = u + at, (also see attached for reference)
Where,
v = final speed (we are asked to find this)
u = initial speed = 0 (because it starts from rest)
t = time taken = 5s
We simply substitute the given values into the equation:
v = u + at
v = 0 + (4)(5)
v = 20 m/s
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
