Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
Answer:
Unbalanced
Explanation:
There is an acceleration. Look at the direction of the arrows.
Answer:
The minumum speed the pail must have at its highest point if no water is to spill from it
= 2.64 m/s
Explanation:
Working with the forces acting on the water in the pail at any point.
The weight of water is always directed downwards.
The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.
And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.
At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.
Net force = W + (normal force)
But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.
At this point of minimum velocity,
Normal force = 0
Net force = W
Net force = centripetal force = (mv²/r)
W = mg
(mv²/r) = mg
r = 0.710 m
g = 9.8 m/s²
v² = gr = 9.8 × 0.71 = 6.958
v = √(6.958) = 2.64 m/s
Hope this Helps!!!
Answer:
A: They maintain stable concentrations.
Explanation:
Hopefully this helps!
Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
