A feather is dropped on the moon from the height of 1.4m. The acceleration of gravity on the moon is 1.67ms-1. Determine the tim
1 answer:
1.3s
Explanation:
Given parameters:
Height = 1.4m
Gravity on moon = 1.67ms⁻¹
Unknown:
Time for feather to fall = ?
Solution:
To solve this problem, we are going to use one of the motion equation that relates time, gravity and height.
H = ut +
Sine the body was dropped from rest, initial velocity is zero;
H = height
u = initial velocity
t = time
g = acceleration due to gravity
since u = 0;
H =
1.4 = x 1.67 x t²
t = 1.3s
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The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.