A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.202
1 answer:
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Answer:
The electric charge, q (in coulomb units) = 5004 C
Given:
The charge stored as printed on NiMH battery, q = 1390 mAh
Solution:
To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:
1 mA =
1 hour =
Also, we know that the rate of flow of charge is electric current, I:
I =
⇒ q = [tex]I\times t[tex] (1)
where
q = electric charge
I = current
t = time taken for flow of current
Using eqn(1), we get:
q = [tex]1390\times 10^{-3}\times 60\times 60[tex]
q = 5004 A-s = 5004 C
Answer:
Part a)
Part b)
Part c)
Part d)
Part e)
Part f)
It is independent of the speed of the object so all part answers will remain the same
Explanation:
Part a)
As we know that Point P is at height 5R while point Q is at height R
so the work done by gravity from P to Q is given as
Part b)
When it reaches to the top of the loop then its final height from ground is
h = 2R
so work done from P to Q is given as
Part c)
Potential energy at P point is given as
Part d)
Potential energy at Q point is given as
Part e)
Potential energy at top point is given as
Part f)
Since all the answer from part a) to part e) depends only upon the position of the object.
So here we can say that it is independent of the speed of the object so all part answers will remain the same