Answer:
Length of the rod is 0.043m
Explanation:
Mass of the bell (M) = 36kg
Distance of the center of mass from pivot = 0.55m
Moment of inertia (I) = 36.0kgm²
Mass of clipper (m) = 2.8kg
Length of the bell to ring = L
The period of the pendulum with small amplitude of oscillation is
T = 2π √(I / mgd)
Where g = acceleration due to gravity = 9.8 m/s²
T = 2π √(36 / 36*9.8*0.55)
T = √(0.1855)
T = 0.43s
The period of the pendulum is 0.43s
To find the length of the Clapper rod for which the bell ring slightly,
T = 2π√(L / g)
T² = 4π²L / g
T² *g = 4π²L
L = (T² * g) / 4π²
L = (9.8* 0.43²) / 4π²
L = 1.7287 / 39.478
L = 0.043m
The length of the Clapper rod for the bell to ring slightly is 0.043m
Answer:
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Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
<span>a) All the 26m should be used to bent into a square to maximize the total area (= 42.25 m2)
b) 1.4136m should be bent into a square and the rest (= 24.586m) should be bent into an equilateral triangle to minimize the total area (= 29.0835m2)</span>
Explanation:
The given data is as follows.
initial speed (u) = 4.0 m/s, mass (m) = 12 kg
Distance (s) = 4.0 m, time (t) = 2.0 sec
First, we will calculate the acceleration as follows.
s =
4 =
a = -2
Now, the final speed will be calculated as follows.
v = u + at
=
= 0
Therefore, change in momentum will be calculated as follows.
= m(v - u)
=
= -48 kg m/s
The negative sign indicates the change in momentum.
Thus, we can conclude that the change in momentum of the box during this time is most nearly 48 kg m/s.