Answer:
Distance= 2.3864m
Explanation:
So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.
In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.
This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.
Mathematically it is calculated by expression:
M= F×d
The moment caused by the first man is:
M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m
The moment caused by the second man must be equal to that caused by the first by which:
M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒
⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m
At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.
Answer:
4.62 N-s
Explanation:
recall that the formula for impulse is given by
Impulse = Force x change in time
in our case, we are given
Force = 14 N
change in time = 0.33s
Simply substituting the above into the equation for impulse, we get
Impulse = Force x change in time
Impulse = 14 x 0.33
= 4.62 N-s
B. is not a validated bu experimentation
Based on the situation above the the work done was 400 Joules. <span>Q = FS cos(theta) is the so-called work function. It's important to learn the work physics; you'll see it over and over in science/physics class. Theta is the angle between the force vector F and the distance vector S. In your problem we assume theta = 0, the two vectors were assumed aligned.</span>
Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.
r^² x Pressure (pa).
Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.
Explanation: