Answer:
idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops
Explanation:
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
Answer:
load (l)=400N
Effort(E)=50N
mechanical advantage (MA)= load ÷Effort
(ma)=400÷50
(ma)=8
Explanation:
I copy pasted from the answer from the same question. Remember to first check if ur question is there
Use KE= 1/2mv^2
So...
50,000=(.5)(1,000)v^2
50,000=500 x v^2
Divide 500 on both sides
100 = v^2
Square root both sides to get rid of v^2
Therefore v = 10 m/s