A cold glass is left sitting outside on a hot day. Soon, water droplets form on the outside of the glass. Describe the events th
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Answer:
, assuming that the gravitational field strength is
.
Explanation:
Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.
By Newton's Second Law, the net force on this block would be . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.
Let denote the mass of this block. It is given that
. The weight of this block would be:
.
Hence, the force that the boy applies on this block would be upward with a magnitude of .
The mechanical work that a force did is equal to the product of:
- the magnitude of the force, and
- the displacement of the object in the direction of the force.
The displacement of this block (upward by ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:
- the magnitude of the force that this boy exerted,
, and
- the displacement of this block in the direction,
.
.
To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.
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,