Explanation:
Fgravity = G*(mass1*mass2)/D².
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
so, if you triple one of the masses, what does that do to our equation ?
Fgravitynew = G*(3*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity
so, the right answer is 3×12 = 36 units.
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
true if you are refering to the desing of the experimnt as it does identify the variable
Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have

Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have

The block's speed is 31.422 cm/s
The particles can undergo small oscillations around x₂.
The given parameters;
- <em>initial energy of the particles = E₁</em>
- <em>final energy of the particles, E₂ = 0.33E₁</em>
The movement of the particles depends on the kinetic energy of the particles.
When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.
However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.
- The maximum position the particle can oscillate is x₅
- The half position the particles can oscillate is x₃
Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.
Thus, we can conclude that the particles can undergo small oscillations around x₂.
Learn more here:brainly.com/question/23910777