1. The problem statement, all variables and given/known data A screw jack has a single start thread of pitch 7mm and a operating handle 800mm long.when raising a load of 750kg the effort required on thev end of the handle is 26N. determine for these operating conditions the following : (a) the mechanical advantage (b) the velocity ratio (c) the efficiency of the machine (d) the law of the machine 2. Relevant equations ma = load/effort vr = 2 x pie x r/p effiency of the machine = ma/vr x 100% law of the machine = E = aw+b 3. The attempt at a solution (a) mechanical advantage = 750 x 9.81/26N = 282.98 (b) the velocity ratio = 2 x pie x 800mm/7mm = 718.07 (c) the effiency of the machine = 282.98/718.07 x 100% = 39.40% (d) the law of the machine this is where i am struggling i no the formula is E = aw+b where a is the velocity ratio and w is the load however what does the b stand for and if i need to caculate this how do i do this please help ......
Answer:
Explanation:
Let assume that spring reaches its maximum compression at a height of zero. The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:
The spring constant is cleared in the expression described above:
The ratio btw number of atoms.
example: water H2O is the compound. No matter how much water you take (a liter, 10 cubic meters... whole swimming pool), the proportion btw hydrogen atoms and oxygen would be always 2 to 1.
Kinetic energy is movement so A
Potential is still so false
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
