Answer:
move the wire loops closer
Explanation:
because the closer t they are the more concentrated the energy is in that specific area
the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.
Answer:
0.25 m.
Explanation:
We'll begin by calculating the spring constant of the spring.
From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:
Force (F) = 0.1 N
Extention (e) = 0.125 m
Spring constant (K) =?
F = Ke
0.1 = K x 0.125
Divide both side by 0.125
K = 0.1/0.125
K = 0.8 N/m
Therefore, the force constant, K of spring is 0.8 N/m
Now, we can obtain the number in gap 1 in the diagram above as follow:
Force (F) = 0.2 N
Spring constant (K) = 0.8 N/m
Extention (e) =..?
F = Ke
0.2 = 0.8 x e
Divide both side by 0.8
e = 0.2/0.8
e = 0.25 m
Therefore, the number that will complete gap 1is 0.25 m.
Answer:
d = 68.18 m
Explanation:
Given that,
Initial velocity, u = 15 m/s
Finally it comes to stop, v = 0
Acceleration, a = -1.65 m/s²
Time, t = 2.5 s
We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :
So, the hayride will cover a distance of 68.18 m.
