You wad up a piece of paper and throw it into the wastebasket. How far will
2 answers:
The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, )
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:
where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s
Substituting,
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Answer:
0.214 m
Explanation:
In order for the bag to levitate and not fall down, the electrostatic force between the bag and the balloon must balance the weight of the bag.
Therefore, we can write:
where
k is the Coulomb constant
is the charge on the balloon
is the charge on the bag
r is the separation betwen the bag and the balloon
is the mass of the bag
is the acceleration due to gravity
Solving for r, we find the distance at which the bag must be held:
Answer:
A) θ = 13.1º , B) E
Explanation:
A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.
In this reference system the only force that we must decompose is the Normal one, let's use trigonometry
sin θ = Nₓ / N
cos θ = / N
Nₓ = N sin θ
Ny = N cos θ
x-axis (radial)
Nₓ = m a
where the acceleration is centripetal
a = v² / R
we substitute
-N sin θ = -m v² / R (1)
the negative sign indicates that the force and acceleration towards the center of the circle
y-axis (Vertical)
Ny - W = 0
N cos θ = mg
N = mg / cos θ
we substitute in 1
mg / cos θ sin θ = m v² / R
g tan θ = v² / R
θ = tan⁻¹ (v² / gR)
we calculate
θ = tan⁻¹ (25² / 9.8 274)
θ = 13.1º
B) when comparing the equations the correct one is E