There are two forces at play:
- The gravitational force acting downward due to the mass of the bucket and the water that it contains.
- The upward force that your hand exerts on the bucket.
If the magnitude of the force your hand exerts on the bucket equals the magnitude of the gravitational force, the bucket is in static equilibrium. That means the bucket is not moving and the forces acting on it balance each other out, making the net force 0.
Having 0 net force means the bucket doesn't undergo any acceleration, or change in motion.
Answer:
A_resulting = 0.2 m
Explanation:
Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.
With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is
A_res = 2A
A_resultant = 2 .01
A_resulting = 0.2 m
Answer:
The mass of the object involved and the value of the gravitational acceleration
Explanation:
- Gravitational potential energy is defined as the energy possessed by an object in a gravitational field due to its position with respect to the ground:
where m is the mass of the object, g is the gravitational acceleration and h is the heigth of the object with respect to the ground.
- Elastic potential energy is defined as the energy possessed by an elastic object and it is given as:
where k is the spring constant of the elastic object, while x is the compression/stretching of the spring with respect to the equilibrium position.
As we can see from the equations, both types of energy depends on the relative position of the object/end of the spring with respect to a certain reference position (h in the first formula, x in the second formula), but gravitational potential energy also depends on m (the mass) and g (the gravitational acceleration) while the elastic energy does not.
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
