Answer:
when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.
when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.
when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.
when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.
Explanation:
Answer:
Professional education as a science has been defined as a field of educational science that studies the growth of a person into a profession and the related problems. It refers to organized education aimed at the knowledge and skills needed in the profession and working life, as well as growing into active citizenship and membership of society. Professional education as a discipline studies vocational training, skills and learning related to the profession and working life.
It enables young people and adults to pursue goal-oriented learning with the aim of acquiring and developing the necessary skills in the profession and creating the conditions for independent professional activity and continuous development in the profession.
Answer:
A.
Explanation:
NEAR THE CENTER OF TECTONIC PLATES.
Answer:
A. The path of least resistance.
Explanation:
ur welcome again ;)
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct