Answer:
Gravity acts to pull the object down.
The object’s inertia carries it forward.
The path of the object is curved.
Explanation:
The motion of a projectile consists of two separate motions:
- A uniform motion along the horizontal direction, where the velocity is constant; since there are no forces along this direction, the velocity does not change, and so the object continues its motion for inertia --> so, the statement "The object’s inertia carries it forward" is true.
- A uniformly accelerated motion along the vertical direction, with a constant downward acceleration (g=9.8 m/s^2, acceleration due to gravity). So, the vertical velocity changes, due to the presence of the gravity that acts to pull the object down.
- As a result of the combination of these two motions, the object follows a curved path (in particular, it is a parabolic path).
Answer:
1,920 Joules
Explanation:
K.E. = 1/2 mv2
so K.E. = 1/2 (60)(8x8) = 1,920 Joules
Answer:
Electrons are not little balls that can fall into the nucleus under electrostatic attraction
Explanation:
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