What is the period of a simple pendulum 47 cm long (a) on the Earth, and ( b) when it is in a freely falling elevator?
1 answer:
Answer:
a)1.37 s
b)∞ ( Infinite)
Explanation:
Given that
L= 47 cm ( 1 m =100 cm)
L= 0.47 m
a)
On the earth :
Acceleration due to gravity = g
We know that time period of the simple pendulum given as
Here
Now by putting the values
T=1.37 s
b)
Free falling elevator :
When elevator is falling freely then
( This is case of weightless motion)
Therefore
T=∞ (Infinite)
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The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so
Friction = 90 N
The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so
Friction = 90 N
Answer:
Explanation:
Given that.
Force acting on the particle,
Position of the particle,
To find,
(a) Torque on the particle about the origin.
(b) The angle between the directions of r and F
Solution,
(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :
So, the torque on the particle about the origin is (32 N-m).
(b) Magnitude of r,
Magnitude of F,
Using dot product formula,
Therefore, this is the required solution.