Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of
mass m1, in meters per square second.
1 answer:
Answer:
a1 = 3.68m/s²
Explanation:
Given values:
Mass of the block placed on the table, m1 = 12.25 kg
Mass of the block hanging vertically, m2 = 7.5 kg
Acceleration due to gravity, g = 9.8 m/s2
Tension in the string is T
Let the acceleration of mass 1 and mass 2 be a1 and a2
a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,
a = m2*g/(m1 + m2)
a = 7.5 x 9.8 / (12.5 + 7.5)
a = 3.68 m/s²
a1 = a2 = 3.68m/s²
a1 is directed to the right and a2 is directed downwards
Below is a diamonds to show the geometrical arrangements of both masses
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Answer:
- The impulse exerted by the ball is 7.2 kg.m/s
- The average force exerted on the ball by the bat is 600 N.
Explanation:
Given;
mass of the baseball, m = 0.144 kg
velocity of the baseball, v₁ = 20 m/s
velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)
The impulse exerted by the ball is calculated as follows;
J = ΔP = mv₁ - mv₂
ΔP = m(v₁ - v₂)
ΔP = 0.144 [20 - (-30)]
ΔP = 0.144 ( 20 + 30 )
ΔP = 0.144 (50)
ΔP = 7.2 kg.m/s
The average force exerted on the ball by the bat is calculated as;
Answer:
(a)
(b)
Solution:
As per the question:
Mass of the object, m = 1.30 kg
Length of the rod, L = 0.780 m
Angular speed,
Now,
(a) To calculate the rotational inertia of the system about the axis of rotation:
Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:
(b) To calculate the applied torque required for the system to rotate at constant speed:
Drag Force, F =