Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
Answer:
80 m/s
Explanation:
Given:
a = -5 m/s²
v = 0 m/s
Δx = 640 m
Find: v₀
v² = v₀² + 2a(x − x₀)
(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)
v₀ = 80 m/s
A. logic, would be your answer i believe!
The amount of heat energy required to raise the temperature of a unit mass of a material to one degree is called D. its heat capacity.
The relationship of the heat when applied to the object and the change in temperature of the object when heat is being applied is directly proportional to each other. This means that when heat is applied to the object, the temperature of the object increases and when heat is not applied to the object, the temperature of the object decreases.
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
