Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
It depends on the mass of an object and acceleration because of the gravity and the height of an object
Answer:
the force between the building and the ball is non-conservative (friction-type force)
Explanation
Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.
When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.
When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall
Consequently, the force between the building and the ball is non-conservative (friction-type force
<h3>In macroeconomic theory,liquidity preference is the demand for money, considered as liquidity.</h3>
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground
