name five influences that guide the decisions that you make.
The correct answer to what additional measurement the student requires is; D. The student has enough information to make the determination
<h3>Understanding Change in Momentum</h3>
Formula for change in momentum is;
ΔP = m(v - u)
Now, this change in momentum is also equal to impulse which has the formula; Impulse = Force(F) × Time(t)
I = F × t
Thus;
m(v - u) = F × t
Now, we are given;
Force exerted on the object and the time interval. Thus;
ΔP = F × t
Thus, the student has enough information to find the change in momentum
The missing options are;
a. The mass of the object.
b. The final speed of the object MOH 5000
c. The distance fallen by the object
d. The student has enough information to make the determination
Read more about Impulse and Change in momentum at; brainly.com/question/20586658
A) The x-component of the shot's acceleration while in flight is; aₓ = 0 m/s²
B) The y-component of the shot's acceleration while in flight is;
a_y = -9.8 m/s²
C) The x-component of the shot's velocity while in flight is; vₓ = 7.55 m/s
D) The y-component of the shot's velocity while in flight is;
a_y = 9.32 m/s
<h3>Projectile motion</h3>
A) The x-component of the shot's acceleration while in flight?
aₓ = 0 m/s²
B) Since the object is thrown vertically, it means that it is against gravity and as such acceleration due to gravity will be negative,
Thus, the y-component of the shot's acceleration while in flight is;
a_y = -9.8 m/s²
C) The x-component of the shot's velocity at the beginning of its trajectory is; v_x = v cos θ
v_x = 12 cos 51
v_x = 7.55 m/s
D) The y-component of the shot's velocity at the beginning of its trajectory is; v_y = v sin θ
v_y = 12 sin 51
v_y = 9.32 m/s
Read more about Projectile Motion at; brainly.com/question/11049671
When the spring is stretched to 11.0 cm - 8.0 cm = 3.0 cm = 0.030 m away from equilibrium, it stores
1/2 (150 N/m) (0.030 m)² = 0.0675 J
of potential energy, while stretching it to 14.0 cm - 8.0 cm = 6.0 cm = 0.060 m from equilibrium stores
1/2 (150 N/m) (0.060 m)² = 0.27 J
Then the change in potential energy is 0.27 J - 0.0675 J = 0.2025 J ≈ 0.21 J
