How does the end point differ from the equivalence point of a titration?
1 answer:
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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).
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:
Centripetal acceleration.
Explanation:
Centripetal acceleration is a property of a body moving in a uniform circular path and it is directed radially towards the center of the circle in which body is rotating.
The force which causes this acceleration is centripetal force which is also directed towards the center of the circle and pulls the body towards its center.
It is calculated through following formula
where v is velocity and r is the radius of the circle.