A ball is thrown vertically downward from the top of a 30.6-m-tall building. The ball passes the top of a window that is 10.7 m
1 answer:
Answer:
v = 19.6 m/s
Explanation:
Height of building = 30.6 m.
Height of window from the ground level= 10.7 m.
Acceleration due to gravity = 9.8
At initial condition ball at rest condition so u= 0 m/s.
Lets take when passes through the window ,velocity is v.
Here acceleration is constant so we can apply motion equation .
We know that
v= u + a t
So by putting the values
v = 0 +9.8 x 2
v = 19.6 m/s
So the velocity of ball is 19.6 m/s when passes through the window after 2 s.
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The drift velocity is given by the equation:
Where
I = current
n = Number of free electrons
A = Cross-Section Area
q = charge of proton
Our values are given by,
The hall voltage is given by
Where
B= Magnetic field
n = number of free electrons
d = distance
e = charge of electron
Then using the formula and replacing,