Newton's three laws of motion can be used to describe the motion of the ice skating.
<h3>Newton's first law of motion</h3>
Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.
- Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.
<h3>Newton's second law of motion</h3>
Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.
- Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.
<h3>Newton's third law of motion</h3>
This law states that action and reaction are equal and opposite.
- Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.
Learn more about Newton's law here: brainly.com/question/3999427
Answer:
The acceleration is 6 [m/s^2]
Explanation:
We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.
![v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2B%20a%2At%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%20%3D%20final%20velocity%20%3D%2022%20%5Bm%2Fs%5D%5C%5Cv_%7Bi%7D%20%3D%20initial%20velocity%20%3D%204%20%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%3D%203%20%5Bs%5D%5C%5C)
Now replacing the values we have:
![a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_%7Bf%7D%20-%20v_%7Bi%7D%20%7D%7Bt%7D%20%5C%5Ca%3D%5Cfrac%7B22%20-%204%20%7D%7B3%7D%5C%5Ca%20%3D%206%20%5Bm%2Fs%5E%7B2%7D%20%5D)
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
His total displacement from his original position is -1 m
We know that total displacement of an object from a position x to a position x', d = final position - initial position.
d = x' - x
If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.
Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.
His final position, x" after moving back 6 m is gotten from
x" - x' = -6 m
x" = -6 + x'
x" = -6 + 5
x" = -1 m
Thus, his total displacement from his original position is
d = final position - initial position
d = x" - x
d = -1 m - 0 m
d = -1 m
So, his total displacement from his original position is -1 m
Learn more about displacement here:
brainly.com/question/17587058
