If we rub two neutral objects
1 answer:
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Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
And the distance traveled downwards is:
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
Replacing into the first equation:
Rationalizing:
Let's call v1 the final speed of the package dropped from a height H. Thus:
Let v2 be the final speed of the package dropped from a height 4H. Thus:
Taking out the square root of 4:
Dividing v2/v1 we can compare the final speeds:
Simplifying:
The final speed of the second package is twice as much as the final speed of the first package.
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As
For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²