<h2><u>Answer:</u></h2><h2>A. Displacement</h2><h2>B. Time</h2><h2 /><h2>Your Welcome ❤</h2>
Answer:
denoting, relating to, or operated by a liquid moving in a confined space under pressure.
Answer:
The necessary information is if the forces acting on the block are in equilibrium
The coefficient of friction is 0.577
Explanation:
Where the forces acting on the object are in equilibrium, we have;
At constant velocity, the net force acting on the particle = 0
However, the frictional force is then given as
F = mg sinθ
Where:
m = Mass of the block
g = Acceleration due to gravity and
θ = Angle of inclination of the slope
F = 5×9.81×sin 30 = 24.525 N
Therefore, the coefficient of friction is given as
24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479
μ × 42.479 N= 24.525 N
∴ μ = 24.525 N ÷ 42.479 N = 0.577
Answer:
∑Fy = 0, because there is no movement, N = m*g*cos (omega)
Explanation:
We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.
If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.